CALIFORNIA STATE UNIVERSITY, NORTHRIDGE
PASCAL'S TRIANGLE:
ITS HISTORY, PATTERNS, AND APPLICATIONS
A thesis submitted in partial satisfaction of the
requirements for the degree of Master of Science in
Mathematics
by
Kathleen Lynn Davidson
May 1988
The Thesis of Kathleen Lynn Davidson is approved:
Barnabas Hughes .
Blattner, Committee Chair
California State University. Northridge
ii
DEDICATION
To my son, Jeremy,
may he someday know and appreciate
the rewards of hard work.
And to Dave Okasaki, who already does.
iii
ACKNOWLEDGMENTS
I would like to express my most sincere and deepest
appreciation to the people in my life who helped me and stood by me
this past year.
To my committee: Dr. John Blattner, Dr. Muriel Wright, and Dr.
Barnabas Hughes, for their invaluable help, extensive time, and
continuous encouragement.
To my colleagues in the math department at Saugus High School:
Jill Iwamoto, Don Steele, and John Pletta, for putting up with me this
past year. They are the best.
To my mother, Jerry, my sister, Becky, and her husband, Cary,
for helping me out on the homefront, making it possible to put in the
many hours needed for this paper.
To my son, Jeremy, for taking the back seat to this project on
many a weekend and evening, and accompanying me on so many
CSUN trips.
To Paul, for the love and support only he could give.
To Dr. Wally Prelle, for his talent and time spent on many of the
graphics and charts in this paper.
To Dave Okasaki, without whom this paper would not exist.
Thank you Dave, for typing this project as if it were your own, for
putting in ten times the number of hours you originally saw, and for
never complaining or giving up. I owe you so much. And to his son,
Mike Okasaki, for the creative graphics he contributed, thank you!
Lastly, to my dear friend and colleague who began this all, Judy
Stevens. We have spent the last five years together teaching, taking
classes, studying, crying, and laughing. Without Judy by my side, I
probably would not be writing an acknowledgment page at all. I never
thought I would say this--"Judy, we made it!"
Love,
Kathie
iv
CONTENTS
Page
DEDICATION.............................................................................................................
ACirnOWIEOOlVIENIS..........................................................................................
iii
ABSIRACI'.................................................................................................................
vii
iv
Chapter
1. BLAISE PASCAL. (1623-1662).............................................................. 2
2. TIIE IllSfORY OF TIIE 1RIANGIE.. .. ... ...... ...... ... .. . ... .. .. .. .. . .... .. .... .... . 8
3. PASCAL'S TREATISE ON THE ARITHMETIC
'IRIAN"GIE.......................................................................................... 15
4. MA.TIIEMA.TICA:L INDUCTION.......................................................... 3 9
5. TI-IE MODERN" DAY1RIANGIE......................................................... 41
//
6. COIVIBINA.TIONS...... ...... ............................ ..... .. ........ .. .......... ... .... .. .. ........ 4 5
7. BINOMIAL COEFFICIENTS AND THEv,
.
pASC::A:L f'YRAMID............ .... .. ... .. .. .. .. .. .. ... .. .. ... .. .. .. ... ... .. .. .. .... ... .. .. ... 5 5
Binomial Coefficients..................................................................... 5 6
The Pascal f'yrarnid....................... ..... .. .. ... .. .. ... .. ... . .. ... .. .. .. .. .. .. ... .. .. 7 0
8. PROBABILI'IY AND THE PROBLEM OF POINTS........................ 8 7
f>robabilit;y.......................... ... ...... .......... ......... .......... ......... .......... ....... 8 8
The f>roblem of Points.................................................................. 9 3
9. PATIERN"S.......\;·..................................................................................... 100
Rows, Columns, and Diagonals................................................ 1 0 1
Hexagons.......................................................................................... 1 13
Fibonacci Seq_uences... ............. .... .. .. ............ .... .. .. .. .. .. ... .. .. ... .. .. .. . 1 16
Primes............................................................................................... 119
Multiples.......................................................................................... 12 4
Figurate Numbers......................................................................... 13 7
Higher Dimensional Figurate Numbers................................ 144
10. .APPUCATIONS.... ............... ........... .... ................ .............. ..... .... ..... .. .... .. 15 2
Simplexes........................................................................................ 1 53
Dividing Space............................................................................... 1 56
Polygons Inscribed in a Circle................................................. 166
Diagonals of Polygons.................................................................. 16 8
v
ABSTRACT
PASCAL'S TRIANGLE:
ITS HISTORY, PATTERNS, AND APPLICATIONS
by
Kathleen Lynn Davidson
Master of Science in Mathematics
A study of the life of Pascal and an examination of the existence
of the triangle prior to Pascal's work is an important preliminary to
the discussion of Pascal's triangle itself. Not only does the history of
the man and the triangle make for interesting reading, it allows the
reader to understand the importance and the role of Pascal's work.
The first two chapters of this paper are devoted to just this.
Pascal's actual treatise is examined and studied in detail followed
by a short chapter on mathematical induction, a process used
extensively in this paper. The modem-day triangle, showing the
triangular array of numbers shifted, is often employed to make use of
the many pattems embedded in the triangle and is so discussed.
The middle section of the paper covers topics that Pascal
himself addressed at the close of his treatise. Four topics, linked
directly to his triangle, make up the theme of this section:
combinations, binomial coefficients, probability, and orders of
numbers.
The final two chapters explore some of the countless patterns,
extensions, and applications of this array of numbers.
vii
)
/
Chapter 1
BlAISE PASCAL (1623-1662)
The next few pages are intended to give insight into the
fantastic genius of Blaise Pascal. WhHe only a few pages make up this
chapter on his biography, Pascal's entire, but short, life is outlined
here. It is important for the reader to know that Pascal had broad and
practical interests. His arithmetical triangle was but one result of this
man's genius.
2
Versatility is not inevitably the companion of genius. It is not
altogether common to find a man who is at the same time a clever
experimental physicist, a creative mathematician, an inventor with an
eye to money-making, a gifted writer whose artistry places him among
the foremost French stylists, and a religious philosopher of singular
originality and ardor. Blaise Pascal was such a man.
Every writer who deals extensively with the history of
mathematics pays considerable attention to this famous man. His
originality and ability were remarkable; had he confined his attention
to mathematics he might have enriched the subject with many more
remarkable discoveries. But after his early youth he devoted most of
his small measure of strength to theological questions, philosophical
thought, and French prose. [8, p. 89]
The Pascals were well-to-do, substantial people of considerable
local importance, and had for generations audited the accounts of the
province of Auvergne, France or presided over its courts. Although the
family could boast a title of nobility granted by Louis XI in recognition
of the faithful services of one Etienne Pascal, an important fiscal
officer in the King's entourage, the title was of parliamentary rather
than of noble condition.
Another Etienne Pascal was President of the Court of Aids, a
court having jurisdiction in all matters pertaining to indirect taxation,
when his second of three children, Blaise, was born- at Clermont
Ferrand in the province of Auvergne, France on June 19, 1623.
Etienne had married in 1618 Antoinette Began, the pious daughter of
a family of the merchant class.
This was a period of scientific discovery and advans~. TheJirst
half of the seventeentlf century was to see" the death of Francis Bacon,
the work of Galileo, Kepler, Descartes, Roberval, and Fermat, the birth
of Newton, Huygens, and Leibniz. But in spite of the spirit of true
scientific thought, there was still a wide-spread faith in astrology,
witchcraft, and omens. [1, p. 180]
Blaise's mother died in 1626 when Pascal was three. Etienne
moved Blaise, his sister, and brother, to Paris in 1631. Etienne, a man
of good mathematical ability, was much worried by Pascal's delicate
constitution and devotion to his books, so he prohibited the boy from
.-..-.......-~-.........,_~.~~..._.~, •. ~,,,--•~-.-,...~,.,-v<r"""'-......,_
-
____ .,.-•'·"•·,,~,,.,.~·.•"~···-
"'•--·~--~
'
,_··'"" ···•·"'
3
·~-"'-,.
.-·-··_..-·
.·
-- .•. ,
thinking about mathematics. Finally, at the age of twelve, Blaise
demanded to know what the subject was about. On learning the
answer from his father he proceeded to devour its contents. The story
has it that by himself he worked out many of the theorems of
. elementary geometry. Some stories report that Blaise reinvented the
·. equivalent of Euclid's first thirty-two theorems. When this precocity
1 was discovered, the prohibition was removed; and the young Pascal
~;
made surprising progress with his studies. Two years later he was
admitted to the weekly scientific meetings of the great French
mathematicians of the time. [26, p. 360]
At sixteen, he composed a work on the conic sections which
built upon the work of Apollonius so skillfully that it won the praise of
Descartes. The work was completely lost and all we have now is a
short fragment called "Essai pour les coniques."
In 1640 the intensity of the geometrical and other intellectual
labors began to affect seriously the health of the frail teenager. From
1640 to 1647 Blaise only occasionally retumed to the intense
intellectual stimulation of Paris. He had already achieved a reputation
in mathematics that was enhanced in 1642, when he designed an
"arithmetic machine" to help his father with computations involved in
tax collecting. The machine, essentially the first digital calculator,
employed the movement of gears to add and subtract. He produced
and/sold the calculating machine, but few could afford it because of its
/
.high price. His contemporaries considered the calculator his greatest
achievement. [7, p. 307]
Religion became especially important to the Pascal family when
they were introduced to Jansenism, a version of Catholicism, in 1646.
However, no behavior was more riddled with contradictions than that
of Pascal. Conflicting beliefs and desires produced strange vagaries of
conduct and caused him to oscillate between the sacred and the
profane. His literary efforts were divided between serious
argumentation on theological controversies and counsel on love.
Deeply disturbed by the differences between the doctrines of the Bible
and the dogma of the Roman Catholic Church, he nevertheless ignored
both when he sought to rob his sister of her inheritance. He awarded
to himself a prize he had offered for competition to the scientists of
4
his time, and then complained of their lack of sincerity in the pursuit
of knowledge. Though he worried about the way to salvation, he
transgressed sufficiently to be in dire need of finding it. His fervid joy
in his religious experiences equaled those of a saint, but his conduct
toward people was marred by the excesses of a sinner. [26, p. 360]
Although Pascal turned his attention to religious and theological
questions after 1646, this did not end his intense research in natural
philosophy and mathematics. He performed his variations on
Torricelli's experiment, which resulted in his "New Experiments
Concerning the Vacuum" in 1647. This in tum led to his investigation
of the action of fluids under pressure of air which established his
reputation as one of the founders of hydrodynamics. His results
appeared in the book, Great Experiments Concerning the Equilibrium
of Fluids (1663).
On the advice of his physician, Pascal put aside some of his
research in 1651. He read a problem posed by his friend the
Chevalier de Mere about the division of stakes in a game of chance
which led Pascal to study the theory of probability. In his
correspondence with Pierre Fermat in 1654 and the writing of his
Traite du triangle arithmetique, he laid the foundations for the
calculus of probability. [7, p. 308)
In 1653 Pascal had to administer his father's estate. He was
contemplating marriage when an accident again turned the current of
his thoughts to a religious life. He was driving a four-in-hand on
November 23, 1654 when the horses ran away. Pascal was saved
miraculously only by the traces breaking. Always somewhat of a
mystic, he considered this a special summons to abandon the world.
He shortly moved to Port Royal, Paris, where he continued to live until
his death. Blaise submitted himself totally to Jesus. During 1656 and
1657 his talents were employed by the Jansenists and he wrote the
Lettres Provinciales, which were directed against the Jesuits, the
enemies of the Jansenists. He wrote these under the pseudonym
Louis de Montalte, for they were published secretly at great risk to
printer and author. [2, p. 281]
To distract his mind from a persistent toothache in 1658, he
began to reflect about the cycloid; after a time he realized that his
5
tooth had ceased to ache. Taking this as a sign of divine permission to
continue, Pascal worked furiously for eight days to solve many
problems of the cycloid. His solutions were later published in a brief
History of the Cycloid and in four Letters written under the
pseudonym Amos Dettonville (an anagram on the name Louis de
Montalte, which Pascal used on the Provincial Letters). Perhaps he
did not want the world to know that he had again engaged in the
science that he had once renounced. [5, p. 426]
The work of Pascal on the cycloid had one by-product of
surprising importance; it served as the immediate inspiration for
Leibniz in his invention of the differential and integral calculus. One of
Pascal's Letters (1658) involved certain calculations that resembled
the evaluation of the definite integral of the sine function. Pascal was,
however, unable to hit upon the crucial point that would give a general
theory of integration. Leibniz later wrote that while he was reading
Pascal's Letter, a light suddenly burst on him and then he realized
what Pascal had not. Perhaps if Pascal had not died so young, or had
he not abandoned mathematics for the religious life, he would have
had the honor accorded Newton and Leibniz for the discovery of the
calculus.
In other work, Pascal believed that he had perfected
Bonaventura Cavalieri's "theory of indivisibles." [5, p. 426]
Mter helping to plan a public transportation project in Paris in
1659, Pascal spent his final years planning a book in Christian
apologetics. Fragments he wrote were discovered at his death and
published under the title Pensees in 1670. The Pensees and ~e
Provencial Letters are read today as models of early French literature.
[13, p. 262]
Constitutionally delicate, Pascal had injured his health by his
incessant study; from the age of seventeen or eighteen he suffered
from insomnia and acute dyspepsia, and at the time of his death was
physically worn out. As death approached, Pascal's life became more
austere·. He gave his possessions to the poor and continually strove for
complete detachment from those he loved. In 1662 he was seized
with a violent illness which lingered for two months. Blaise Pascal
died on August 19, 1662 at the age of 39. [2, p. 283]
6
His health was such that most of his life was spent racked with
physical pain. and from early manhood he also suffered the mental
torments of a religious neurotic. Pascal's mathematical reputation
rests more on what he might have done than on what he actually
effected. It is very seldom disputed that Blaise Pascal was the greatest
"might-have-been" in the history of mathematics. [2. p. 281]
7
Chapter 2
THE HISTORY OF THE TRIANGLE
Many believe that Pascal's triangle originated with Pascal
himself. History shows that the triangle preceded Pascal by at least
1800 years. Men before Pascal employed the'triangle to varying
degrees, but it was Pascal who first studied it in depth and thus his
name will be forever linked to this array of numbers. The following
pages show the occurrence of the triangle prior to Pascal.
8
Most people enjoy poetry. Many have tried writing it. Few have
been proficient. Fewer still have actually analyzed its structure. One of
the earliest to systematize meter for verse was the Hindu poet Pingala
(circa 200 B. C.).
The problem he set himself to solve is this: how many different
meters can be produced with words of any number, n, of syllables? He
figured quite correctly that monosyllabic words consist of either short
or long syllables. Disyllabic words consist of all long syllables or all
short or two arrangements of long and short syllables. Words of three
syllables have four forms: all short, three arrangements of two long
and one short, three arrangements of two short and one long, and all
long. And so on. He made a scheme of his system which he called
"meru prastara":
= 20 = 1
1
1
2
1
1
1
3
4
= 21 =2
1
4
=4
= 23 =8
1
3
6
= 22
1
= 24 = 16
1
He noted the total number of arrangements in each horizontal row to
be 2n, n being the number of syllables. He further observed that every
element in the horizontal rows is the sum of the two numbers
immediately above it, such as 1 + 3 = 4. Pingala left us the earliest
record of this remarkable arrangement of numbers.
Chinese references indicate that the "arithmetic triangle"
appeared in a work of Chia Hsien about 1050. An arrangement of
binomial coefficients was known to the Arabs about the same time that
it was being used in China. Omar Khayyam (circa 1050-1130) referred
9
in On Demonstrations of Problems of Algebra and Almucabola to a lost
work in which he dealt with the arithmetic triangle. Another Persian
mathematician, Al-Tusi (circa 1200-1275), in a work called Collection
on Arithmetic by Means of Board and Dust, used the binomial
expansion and furnished a table of binomial coefficients in triangular
form up to the 12th power. Thus the so-called Pascal triangle is in
reality the product of a much earlier Eastern culture.
[5, p. 429]
During the Sung dynasty a Chinese mathematician, Yang Hui, left
us with the earliest extant presentation of the arithmetic triangle.
The book containing this is dated 1261. The triangle is again found in
a later book written by Chu Shi-kie in 1303. A triangular arrangement
of binomial coefficients through the eighth power, written in rod
numerals and a round zero sign, is shown in figure 1. It is found in his
treatise The Precious Mirror of the Four Elements. Chu speaks of the
triangle as already ancient in his time. It would appear, then, that the
binomial theorem was known in China for a long time. We again see
the properties of the binomial coefficients discussed by the Persian
mathematician Jamshid Al-Kashi in his Key to Arithmetic of c. 1425.
[13, p. 174]
Figure 2 shows the first triangular arrangement of the binomial
coefficients to be printed in European books. It appeared in 1527 on
the title page of the Rechnung by Peter Apianus (1495-1552). The
arithmetic triangle seems to have been discovered almost
simultaneously by several authors of the 1500's. Michael Stifel (14861567) in his Arithmetica Integra (1544) calculated the device as far as
the 17th line. Other schematic arrangements for the binomial
coefficients were given by Tartaglia in the Tratlato Generale (1556)
and Cardano in Opus Novum de Proportionibus (1570). Tartaglia gave
the numbers in the triangle through the eighth line, claiming the idea
as his own invention. Cardano presented a table to seventeen lines,
~l::iting Stifel as the original discoverer. [5, p. 429]
Although Blaise Pascal was not the originator of the arithmetic
triangle, being nearly the last of a long line of "discoverers," his name
is forever linked with the triangle because he was the first to make any
sort of systematic study of the relations it exhibited. The merits of
10
{.
r~S3
12to I
-·-~r=-
Pascal's work in this regard are enough to justify the use of his name.
The so-called Pascal triangle appears in his treatise under the title
"Traite du triangle arithmetique, avec quelques autres petits traiN~s
sur la meme maniere." It was written in 1653 and published
posthumously in 1665. In his treatise he provided a method for
finding binomial coefficients, stressed reasoning by recursion, and
correctly stated a formula for finding the number of combinations of n
things taken rat a time.
11
.•
\ J \ ' ~~
~~ ~~ ~~ '~"~
'
'
~ !\'
,If-~ 1;~~ ~ ..~
~
'\'
Figure 1. Pascal's Arithmetical Triangle as Depicted in 1303 by
Chu Shi-kie. [Source: 13, p. 176]
12
JE:ynllaewe
~nno
:\1olge{triinore
t1nber~evtung aHer 1\auflhtan~
l'cd1,
nung in oze'fen Glid)erntmie fchonen ~e
geln ~fi frag(tucfcn 6t!friffcn. eunl)cr~
was fo:rl VJ1 nti 6e~tttbi~~~it in bcr
lien
Wclfche p:ac.tic-a~fi ~ollttn geSzaud~t
Wt\'bt 1 beG glc~d)en
fiinu(llfi tt'iOcr iu
il:eti13(chcr nod) iu Wclfo>er O.nach nt( '
getn(icfc. ~nrd) petrn•n 2Cpiaufi
~on ~~'9/ftJicf/o 2!flronomei
Jngol(fat: <Ort>ina~
,u~~~~ rili1 verfcrtigc~.
""
:}.
.-,
.r~·.,JI
Figure 2. Title-Page of the Arithmetic of Peter Apianus, lngolstadt,
1527, More than a Century Before Pascal Investigated the Properties
of the Triangle. [Source: 35, p. 509]
13
Historic Appearances of the Triangle
c. 200B.C.
Pingala (Hindu Poet)
Table for arranging n words into all possible
meters
c. 1050
Chia Hsien (China)
Referred to in later works
c. 1050-1130
Omar Khayyam (Persia)
On Demonstrations of Problems of Algebra and
Almucabala
c. 1200-1275
AI-Tusi (Persia)
Collection on Arithmetic by Means of Board
and Dust
Fumished a table of binomial coefficients to
the 12th power.
1261
Yang Hui (China)
Earliest extant presentation of the triangle
1303
Chu Shi-kie (China)
The Precious Mirror of the Four Elements
Fumished a table of binomial coefficients to
the sth power
c. 1425
Jamshid Al-Kashi (Persia)
Key to Arithmetic
1527
Peter Apianus (Germany)
Rechnung
1544
Michael Stifel (Germany)
Arithmetica Integra
1556
Tartaglia (Italy)
Tratlato Generale
1570
Cardano (Italy)
Opus Novum de Proportionibus
1654
Blaise Pascal (France)
Traite du Triangle Arithmetique
14
Chapter 3
PASCAL'S TREATISE ON THE ARITHMETICAL TRIANGLE
This section deals with the actual 'Treatise on the Arithmetical
Triangle," which appears in the fifth volume of Oeuvres de Pascal.
While the original is of course in French, I have taken an English
translation, and, although I have used different notation and different
wording to make for easier reading, the following is in the same order,
and of the same flavor as that of Pascal's original.
15
~
I
Pascal's Arithmetical Triangle
exponents of the columns
1
2
3
1
2
3
4
l:S
c
'-A
D
1, 1
1
1
F '-G
I
H
1
2, 3
4
['M
K
L
N
6, 10
1
3
5
E.
8
9
10
1
1
1
1
1
6
7
8
9
1t
21
28
36
20, 35
56
84
R
s
10
1"--T
u
v
4
1
~
1~
3~
'-Z
6
1
6
21
56
126
7
1
7
28
84
8
1
8
36
9
1
9
10
1
5
7
1
J
5
~
4
6
70-........ 126
Pascal calls his "arithmetical triangle" a figure constructed as
follows:
From any point, A, two lines are drawn perpendicular to each
other, AE, AV, in each of which are taken as many equal and
consecutive parts as desired. Beginning with A the rows and columns
are each numbered 1, 2, 3, 4, etc. along the left side and the top of
the arithmetical triangle. These numbers are called the "exponents"
of the sections of the lines.
Next, the points of the first section in each of the two lines is
connected by a line, which is called the "base" of the resulting triangle.
In the same way the two points of the second section are
connected by a line, making a second triangle of which it is the base.
And in this way connecting all the points of sections with the
same exponent, as many triangles and bases are constructed as there
are exponents.
Through each of the points of section and parallel to the sides
lines are drawn whose intersections make little squares called "cells."
16
I
.
Cells between two parallels drawn from left to right are called
"cells of the same parallel row," as, for example, cells A, B, C, etc., or
F, G, H, etc.
Cells between two lines drawn from top to bottom are called
"cells of the same perpendicular row," as, for example, cells A, F, K,
etc., orB, G, L, etc.
Those cut diagonally by the same base are called "cells of the
same base," as, for example, F, B, or K, G, C, or Q, L, H, D.
Cells of the same base equidistant from its extremities are called
"reciprocals," as, for example, L, H, and R, I, because the parallel
exponent of one is the same as the perpendicular exponent of the
other. In the above example R is in the second perpendicular row and
in the fourth parallel row and its reciprocal, I, is in the second parallel
row and in the fourth perpendicular row, reciprocally. Cells with
exponents reciprocally the same are in the same base and are
equidistant from its extremities.
Note that the perpendicular exponent of any cell when added to
its parallel exponent is equal to the exponent of the base plus one.
Note that each base has one more cell than the preceding base,
and that each has as many cells as its exponent has units: thus the
second base, FB, has two cells, the third, KGC, has three cells, etc.
The number of the first cell A, which is at the right angle, is
arbitrary; but that number having been assigned, all the rest are
determined, and for this reason it is called the "generator" of the
triangle.
Pascal then gives the rule to obtain the number of any given cell.
This will be used to prove several of the consequences that follow.
Pascal's rule: the number of each cell is equal to the sum of the
number of the cell immediately to the left and the number of the
cell immediately above.
17
For example:
11111111
cGJ:
1o~r
4
1 4 1cf20\ 35
3 = 1+ 2
21=15+6
35 = 15 + 20
15~
Pascal proceeds to prove nineteen of what he calls the most
important consequences of the arithmetical triangle. He considers
triangles generated by unity, i.e. A = 1, but the consequences will hold
for all others.
Again, I have changed some of the notation and wording for
easier reading. I have also, in many cases, proved the consequences a
little more thoroughly or convincingly, if you will, than Pascal did.
However, the proofs remain in the spirit of Pascal. One must
recognize that in the notation of the day, "general" proofs could only
be given by means of representative cases. Pascal knew that his proofs
worked for all the cells besides the cells cited in his proofs.
Parallel rows will be called "rows" and perpendicular rows will
be called "columns" in the following consequences.
Consequences # 1-11 are statements of equalities.
1. Each cell of the first row and each cell of the first column are the
same as the generating cell, A. i.e. all the cells in the first row and the
first column are the same. (in our triangle these numbers are all one
since A= 1.) These cells are the "extremes."
Proof: Each cell in the first column has no cell to the left, so we can
let zero be the number to the left. Each cell in the first row
has no cell above, so we can let zero be the number above.
Consider the first column:
F=O+A=A
}
K = 0 + F = F =A
and so on.
18
rule
Consider the first row:
B=A+O=A
}
C = B + 0 = B =A
rule
and so on.
2. Each cell is equal to the sum of all the cells of the row above from
the same column of the given cell to the first columh, inclusive.
Proof: Consider any cell, say N.
Now,
N=I+M
}
=I+H+L
=I+H+G+K
rule
consequence 1
=I+ H + G + F
Check:
F
1
G
2
H
3
I
4
N
10
N =I+H+G+F
10 = 4 + 3 + 2 + 1
and so on.
3. Each cell is equal to the sum of all the cells of the column to the
left from the same row of the given cell to the first row, inclusive.
Proof: Consider any cell, say N.
Now,
N=M+I
=M+H+D
}
rule
consequence 1
=M+H+C
19
Check:
c
1
H
3
M
N
6
10
N =M+H+C
10 = 6 + 3 + 1
and so on.
4. Each cell is equal to the sum of all the cells in all the rows above
that are in columns to the left of the given cell plus the generator A.
Proof: Consider any cell, say N.
N=C+H+M
=C+H+B+G+L
=C+H+B+G+A+F+K
}
=C+H+B+G+A+F+A
= (A + B + C + F + G + H) + A
Check:
A
1
B
1
c
F
1
G
2
H
3
1
N
10
N =(A+B+C+F+G+H)+A
10 = ( 1 + 1 + 1 + 1 + 2 + 3) + 1
and so on.
20
consequence 3
consequence 1
5. Each cell is equal to its reciprocal. (The reciprocal of the cell in
column a and row b is the cell in column b and row a.)
Proof: Consider the second base, FB.
= B
consequence 1
F
J..
J..
col 1
col2
row2
row 1
Consider the third base, KGC.
K =
C
consequence 1
J..
J..
col 1
col 3
row3
row 1
Consider the fourth base, QLHD.
Q =
D
consequence 1
J..
J..
col 1
col4
row4
row 1
Now,
L =
K+G
rule
L
C+G
consequence 1
=
L
H
rule
=
J..
J..
col2
col3
row3
row2
Similarly it can be shown for all other bases because the
extremes are always equal and the rest can always be
considered as the sum of cells in the preceding base which are
themselves reciprocals.
21
6. Each row is made up of the same sequence of cells as the column
with the same exponent. (i.e. row 1 matches column 1, row 2
matches column 2, etc.)
Proof: Cells in any given row are the reciprocals of the cells of the
corresponding column and are therefore equal by
consequence 5.
7. The sum of the cells in each base is twice the sum of the cells in
the preceding base.
Proof: Consider any base, say the fourth base, QLHD.
Q+L+H+D
=
K+L+H+C
consequence 1
=
K+K+G+G+C+C
rule
=
2K + 2G + 2C
=
2(K + G +C)
Check:
C/
vl
Q/
D/
/1
H/
l/2 l/3
K/
baseS
base 4
/1
L/
vs
Q/
/1
Q + L + H + D = 2(K + G + C)
1 + 3 + 3 + 1 = 2(1 + 2 + 1)
and so on.
22
Q
8. The sum of the cells of each base is equal to A· 2n- 1 where n is the
exponent of the base. (Since A = 1 in our triangle, we shall use 2n - 1 .)
Proof: Consider the first base A:
A=1=2°=2 1 - 1
Consider the second base FB:
F + B = 1 + 1 = 2 = 2 1 = 22 - 1
Consider the third base KGC:
K + G + C = 1 + 2 + 1 = 4 = 22 = 23 - 1
Consider the fourth base QLHD:
4
Q + L + H + D = 1 + 3 + 3 + 1 = 8 = 23 = 2 - 1
and so on to infinity.
9. The sum of the cells of each base is equal to the sum of the cells of
all the preceding bases plus A. (We shall use A= 1)
Proof: We use the sum formula for a geometric series:
a-arn-1
Sn = a+ ar + ar2 + ar3 + ... + arn-2
1 _r , r
We take a= 1, r = 2 to obtain:
·
n-2
1 - 2n-1
1+2+4+8+ ... +2
= 1-2
= -1 + 2n-1
or 2n- 1 = (1 + 2 + 4 + 8 + ... + 2n- 2 ) + 1
sum of the cells
of base n from
consequence 8
sum of the cells of
base 1 through
base n - 1, inclusive,
from consequence 8
23
A
*1
•
Check:
ba.se 1
base 4
base 3
base 2 base 1 generator
Q+L+H+D=K+G+C + F+B + A + A
1+3+3 +1=1+2+1 + 1+1 +
1
+
1
10. The sum of any number of consecutive cells of any base, beginning
at an extremity, is equal to the sum of the same number of cells in the
preceding base, starting at an extremity, plus the sum of the same
cells without the last cell.
Proof: Consider any base, say the fourth, QLHD. Consider taking the
sum of say the first three, Q + L + H.
Now, Q + L + H
consequence 1
=K+L+H
rule
= K+K+G+G+C
= (K + G + C) + (K + G)
24
Check:
c/ D/
/1
Q/
base e
base 4
/1
H/
/2 /3
K/ L/
l/1 /3
Q/
l/1
Q + L + H = (K + G + C) + (K + G)
1 + 3 + 3 = (1 + 2 + 1) + (1 + 2)
and so on.
11. Each cell of the bisector is double the cell above it and is double
the cell to the left of it. (Cells of the bisector are those diagonally cut
by the bisector of the right angle such as A, G, M, T, Z, etc.)
Proof: Consider any cell on the bisector, say M.
M=H+L
rule
=H+H
consequence 5 (since L and H are
reciprocal cells)
= 2 H
(twice the cell above M)
And again,
M=H+L
rule
= L+ L
consequence 5 (since L and H are
reciprocal cells)
= 2L
(twice the cell to the left of M)
25
Check:
H
3
L
3
M
=
2H
=
M
6
2L
6 =2·3=2·3
and so on.
Consequences #13 - 19 are statements of proportions, of which #12 is
the foundation.
12. For any two consecutive cells in a base:
upper cell of the two cells
lower cell of the two cells
=
the number of cells from the upper cell to the upper
extremity of the base, inclusive
the number of cells from the lower cell to the lower
extremity of the base, inclusive
Proof: Pascal proves this consequence by supposing two lemmas:
Lemma 1: For the second base, FB:
B
F
1
the number of cells from B to B, inclusive
= T = the number of cells from F to F, inclusive
Lemma 2: If this proportion is found in any base, it will.
necessarily be found in the following base. (Even though Pascal
did not use the term induction here, we recognize the logical
process as such and we address this issue later in the paper.)
From these two lemmas it is clear that it will be true in all
bases because it is true in the second base by the first lemma;
26
hence by the second lemma it is true in the third base,
therefore in the fourth base, and so on to infinity.
It is therefore necessary to demonstrate only the second
lemma as follows: let this proportion be true in an arbitrary
base, as in the fourth one, QLHD, that is, if
L
3
the number of cells from L to D, inclusive
Q - T = the number of cells from Q to Q, inclusive ' and
H
3
r=3
= 2 = the number of cells from H to D, inclusive
2
d
the number of cells from L to Q, inclusive ·an
D
1
the number of cells from D to D, inclusive
H = 3 = the number of cells from H to Q, inclusive'
then we must show that the proportion holds for the fifth base,
VRMIE. We consider cells M and R.
L
Q
R
L
=
3
1
from the hypothesis
=
L+Q
L
rule
=
3 + 1
3
4
=
3
In the same way,
H
L
M
L
2
2
=
3
3
=
L+H
L
=
2+2
2
=
from the hypothesis
rule
27
4
=
2
=
2
1
By compounding the ratios,
M
L
R
2
=
L
M
or R
1
4
3
6
=4
3
= 2
3
the number of cells from M to E, inclusive
where 2 = the number of cells from R to V, inclusive
The proof is the same for all other bases, since it requires only
that the proportion be found in the preceding base, and that
each cell be equal to the sum of the cell to the left and the cell
above, which is true in all cases by Pascal's rule.'
Check:
E/
/1
I /
M/
/
/4
l/6
R/
l/4
V/
base 5
E
1
I - 4
/1
=
the number of cells from E to E, inclusive
the number of cells from I to V, inclusive
28
I
4
2
M- 6 = 3
=
the number of cells from I to E, inclusive
the number of cells from M to V, inclusive
M
6
3
R =4 = 2 =
the number of cells from M to E, inclusive
the number of cells from R to V, inclusive
R
the number of cells from R to E, inclusive
the number of cells from V to V, inclusive
v-
4
=
1
13. For any two consecutive cells in a column:
the lower cell = the exponent of the base of the upper cell
the exponent of the row of the upper cell
the upper cell
Proof:
Consider any column, say the third, and take any two
consecutive cells in it, say S and M.
M
3
Now, R = 2
M+R
M
=
by consequence 12
rule
3+2
3
5
=3
= the exponent of the base of M
the exponent of the row of M
29
Check:
/
/
/
M/
v
v6
Row3
/
1/
s
10
/
base 5
/
S
1 0 = 5 = the exponent of the base of M
3
the exponent of the row of M
M = 6
and so on.
14. For any two consecutive cells in
the greater cell
the lesser cell
=
a row:
the exponent of the base of the lesser cell
the exponent of the column of the lesser cell
Proof: Consider any row, say the fourth, and take any two consecutive
cells in it, say RandS.
M
3
Now, R = 2
s
R=
=
by consequence 12
M+R
R
rule
3+2
2
5
=2
=
the exponent of the base of R
the exponent of the column of R
30
Check:
Col2
/
/
/
/
/
R/
/
s
10
l/_4
/
base 5 l(_
s
R
10
= 4=
5
2
the exponent of the base of R
the exponent of the column of R
=
and so on.
15. For any arithmetical triangle:
the sum of the cells of any row in the triangle
the last cell of the row
=
the exponent of the base of the triangle
the exponent of the row
Proof: Consider any triangle, say the fourth, AQD, and any row in the
triangle, say the second, FGH.
Now,
F+G+H
H
=
=
=
M
H
consequence 2
4
consequence 13
2
the exponent of the base of the triangle
the exponent of the row
31
Check:
A
B
c
F
G
1
2
H/
/3
K
L/
D/
/
Row2
g/
base 4
/
l/
For the fourth triangle AQD and the second row:
F+G+H
H
=
1 +2+3 _ 6 _ 4 _ the exponent of the base of the triangle
3
- 3 - 2 the exponent of the row
and so on.
16. For any arithmetical triangle:
the sum of any row
_
the exponent of the lower row
the sum of the next row - the number of cells in the lower row
Proof: Consider any triangle, say the fifth, AVE, and any row in the
triangle, say the third, KLM.
Now,
K+L+M
Q+R
s
= w
=
=
consequence 2
4
consequence 12
2
the exponent of the lower row
the number of cells in the lower row
32
Check:
A
RowS
B
D
H
I /
F
G
K
M/
L
3 l/6
1
E/
v
v
R/
Q
1
Row4
c
l/4
V/
base5
v
For the fifth triangle, AVE, and the third row, KLM:
K+L+M
Q+R =
1 +3+6 _ 1 0 _ 4 _
the exponent of the lower row
1 +4 - 5 - 2 - the number of cells in the lower row
and so on.
17. For any cell:
the sum of the cell and all the cells above it
the sum of the cell and all the cells to the left of it
=
the number of cells taken in the column
the number of cells taken in the row
Proof: Consider any cell, say L.
Now,
L+G+B
L+K
=
M
L+K
consequence 3
M
= R
consequence 2
33
=
=
3
consequence12
2
the number of cells taken in the column
. the number of cells taken in the row
Check:
B
1
G
2
K
1
L+G+B
L+K
=
M
6
0?
3+2+ 1
3+ 1
6
3
=4 =2 =
the number of cells in the column
the number of cells in the row
and so on.
18. In every arithmetical triangle, for any two rows equidistant from
the extremities:
the sum of the cells of the upper row
the sum of the cells of the lower row
= the number of cells of the upper row
the number of cells of the lower row
Proof: Consider any triangle, say the fifth, AVE. Consider the second
and fourth rows, FGHI and QR.
B+G+L+R
F+G+H+I
=
Q+R
Q+R.
34
consequence 6
=
the number of cells taken in the column
the number of cells taken in the row
consequence 17
using R as
the cell
= the number of cells taken in the upper row
the number of cells taken in the lower row
Check:
A
B
c
D
H
3
I /
/4
1
Row2
F
G
1
2
K
L
3
Q
R/
/4
1
Row4
E/
/
/
L
V/
base 5
l/
F+G+H+I
Q+R
=
1+2+3+4
1+4
the number of cells in row 2
= the number of cells in row 4
and so on.
19. For any two consecutive cells on the bisector:
~~t_h_e_l-:-::o_w_e_r_c_e_ll_-:-:-
4 times the upper cell
=
the exponent of the base of the upper cell
the exponent of the base of the upper cell plus 1
Proof: Consider cells T and M on the bisector.
Now,
T
4M =
2N
4·2H
consequence 11
35
=
N
4H
5
3
4H
M
N
=
=
=
M
4H
M
=
consequence 14
5
3
consequence 13
~
4·4
5
3
=
2
5
=
6
exponent of the base of M
exponent of the base of M plus 1
1
Check:
/
['-.
/
"
"'
/
/
"
"M/
/6"
/
/
base 5
T
4M
=
"-T
20,
1/
~
I/
20
4·6
"
=
5
6
=
bisector
exponent of the base of M
exponent of the base of M plus 1
and so on.
It is interesting to note that this consequence states that the numbers
on the bisector approach a geometric sequence with a common ratio
equal to four.
36
Pascal ends with the following problem: Given the column and
row of a cell, find the number of the cell without using the
arithmetical triangle.
We will use the example of finding the number of cell I of the
fourth column and of the second row.
Having taken all the numbers that precede the exponent of the
column, that is, 1, 2, 3, take as many natural numbers beginning with
the exponent of the row, that is, 2, 3, 4.
We multiply the first set of numbers: 1 · 2 · 3 = 6 and we
multiply the second set of numbers: 2 · 3 · 4 = 24. The number of cell
I is equal to the quotient of the two products, i. e.
I=2_·3_·_4
1·2·3
Proof: Consider the fifth base VRMIE.
I
I
I = 1 = v
=
M
R
I
M
2
= 3
=
consequence 1
3
2
R
v
4
1
consequence 12
4
Check: U in the fifth column and fourth row:
u = 4 . 5 . 6 . 7 = 35
1·2·3·4
X in the third column and fifth row:
X =
5
.
6
1. 2
= 15
and so on.
37
Pascal continues his treatise on the arithmetical triangle by
discussing four of its uses:
(1) Combinations,
(2) Powers of binomials,
(3) The Problem of Points (probability), and
(4) Orders of numbers.
These topics are discussed in the remainder of this paper.
38
Chapter 4
MATHEMATICAL INDUCTION
It is now appropriate to insert a short chapter on mathematical
induction. Not only have we seen that Pascal put the process to
serious use, but the process of induction will be used extensively in
the remainder of this paper.
39
A theorem provable by mathematical induction involves a
statement about an integer, n, which is to be proved true for all values
of the integer. The proof is in two parts. The first part proves that
the theorem is true for the lowest meaningful value of n, say n = N0 .
The second part of the proof is what has been call the argument from
N to N + 1. One assumes the theorem works for N and shows that, if
it does, then it works for N + 1. This "domino effect" proves the
theorem true for all integers greater than or equal to N0 .
The method of mathematical induction was recognized explicitly
by Maurolico in his "Arithmetica" in 1575, but Blaise Pascal was the
first to appreciate it fully, and he used it extensively in connection
with his triangle, in the calculation of powers of binomials, and in the
application of the triangle to the theory of probabilities.
[9, p. 376]
Consequence 12 of the "Arithmetic Triangle" is of historical
importance because, as we have seen, its proof involves the
recognition of the logical process of mathematical induction and
brought it into the domain of the working mathematician. While the
induction proofs of Maurolico were presented in a somewhat sketchy
style, Pascal followed more nearly modem lines. [5, p. 435]
Maurolico and Pascal had no special designation for the
technique of reasoning by recurrence. The earliest mathematician
who appeared in the fixing of a name to this process of argumentation
was John Wallis who used the phrase "per modum inductionis" in
1656. Augustus De Morga.'"1 suggested the nmne "successive induction"
in 1838 and also referred to the process as "mathematical induction,"
which is the first use of the term. The expression "complete
induction" was used in Germany after Dedekind used it in 1887. In
the present century the name "mathematical induction" has gained
complete ascendancy over other descriptive terms. [5, p. 440]
40
Chapter 5
THE MODERN DAY TRIANGLE
The traditional form of Pascal's triangle has often been modified
in modern texts. The purpose of this is to offer greater clarity in
identifying many patterns. The next few pages look at this change.
41
Pascal's triangle is often seen as a triangular array of natural
numbers differing from the triangle examined by Pascal himself by a
rotation through 45°. The triangular array is symmetrical about a
vertical line from its top number. Pascal's parallel rows become
diagonals and are numbered as shown:
1
Columns are numbered similarly:
1
Note that c 0 = do. c 1 =
d~o
etc.
Numbering begins with 0 instead of 1 to simplify explaining
some of the triangle's basic properties.
42
The horizontal rows are numbered as follows:
rO
r1
1
-1-1
1
r2
1
1
2
1
4
1
4
6
1
r4
10
10
1
5
5
1
r5
1 6 15 20 15 6 1
r6
1 7 21 35 35 21 7 1
1 8 28 56 70 56 28 8 1
r3
3
3
A number is referred to as an element. 1 is the Oth element in
each row, in each diagonal and in each column.
2 is the first element in the second row and in the first diagonal
and column.
4 is the first element in the fourth row and in the third diagonal
and column.
6 is the second element in the fourth row and in the second
diagonal and column.
The number of elements in each row is finite. (The nth row
contains n + 1 elements.) The number of elements in each diagonal
and in each column is infinite since they never end.
The arithmetical triangle is now constructed by the following
definition (Pascal's rule again, as defined in Chapter 3): each number
is the sum of the two numbers directly above it.
1
1
1
6
1
7 21 35 35 21 7 1
8 28 56 70 56 28 8 1
1
1
1
The triangle can at any stage be enlarged by affixing the number
1 at each of the right- and left-hand extremities of each new row. The
other numbers are found by using the above definition. Thus the
triangle continues infinitely.
-----
.-
43
The triangle below is computed through the 22nd row.
[Blank grid. Source: 34, p. 76]
44
Chapter 6
COMBINATIONS
We stated at the end of chapter 3 that Pascal continued his
treatise on the arithmetical triangle by discussing four of its uses:
combinations, powers of binomials, the problem of points, and orders
of numbers. In this chapter we will explore Pascal's first topic,
combinations. We will see how this is used in set ~-~~ry andin~)~!ter
chapter how it is used in _p~~-~~-~!!i:!Y·
45
In his discussions on probability, Pascal often deals with
combinations. He states: "The word 'combination' has been used in
several different senses, so that to avoid ambiguity I am obliged to say
how I understand it. When of many things we may choose a certain
number, all the ways of taking as many as we are allowed out of all
those offered to our choice are here called the 'different
combinations'."
So a combination is any collection of r members chosen out of a
set of n things or a "combination of n things taken r at a time." It is
important to note that when listing combinations the order of the
elements is not important. That is, abc and hac are not considered
different combinations.
For example, suppose we have four cards, one marked "A", one
marked "B", one marked "C", one marked "D", and we want to
combine them in groups of two. There are six possibilities: AB, AC,
AD, BC, BD, CD. We say that these are the combinations of four things
taken two at a time, or using modern notation:
(24)
= 6. (The
notation (~) is from Euler.)
We consider all the combinations given four choices and use the
notation (~) to represent r elements taken from n elements:
(~) =
1
one way to take none at a time
(i)
=4
four ways to take one at a time (A, B. C, D)
(i)
=6
six ways to take two at a time (AB, AC, AD, BC, BD,
CD)
(~) = 4
four ways to take three at a time (ABC, ABD, BCD)
t'~ib
46
(!)
one way to take four at a time (ABCD)
= 1
We note that 0:::;; r : :; n for (~) to have meaning since r
represents the number of elements taken out of a set of n elements.
We see that 1 4 6 4 1 is the fourth row of Pascal's triangle, i.e.
the fourth row may be written as:
(6)
(i)
(~)
(~)
(!).
In fact, upon inspection, we see that we may write the zeroth
row through the fifth row as:
We see since all the rows start and end with the number 1, that
(~) = (~) =
1.
It is appropriate to call this relation the "boundary condition" of
the triangle. This agrees with our combinatorial usage of the symbol
(~) since there is always only one way to choose no elements and
only one way to choose n elements out of a set of n elements.
47
If we use the symbol
(~) to represent the rth element in the
nth row of Pascal's triangle as well as the number of ways to take r
elements from n elements, we have a direct relationship between the
triangle and combinatorial operations.
The equation
serves a twofold purpose. It is a recursion formula for the elements in
the triangle as well as for combinations. We prove this equation for
combinations.
Proof:
Suppose we have a set of n + 1 elements and we want to
count the number of ways to take r elements, i.e.
( n+r
1)·
Let "a" be one of then+ 1 elements. Then the number of
ways to take r elements will be equal to the sum of the
number of ways to take r elements with "a" included and
the number of ways to take r elements with "a" not
included.
If "a" is among the r elements taken then r - 1 elements
remain to be taken from the n elements left and the
number of way~ to do this is
(r~1).
If "a" is not among the r elements taken then r elements
still remain to be taken from a choice of n elements (since
"a" is not a choice.) The number of ways to do this is(~)
Therefore, we have
+
48
And now that we are using (~) to represent elements in the
triangle, we see that the above equation duplicates Pascal's rule which
we used previously: any element of the triangle is the sum of the two
elements directly above it. The equation tells us that the rth element
in the (n + 1)st row is equal to the (r - 1)st and the rth elements in
the nth row, the row directly above the (n + 1)st row.
Therefore, from any row we may obtain the following row. In
general, we have:
1)
n +
( n+1
We see that it is appropriate to call the last formula for Pascal's
rule the "recursion formula" of the Pascal triangle since each element
depends on two elements from the previous row.
Often it is desirable to compute an element in Pascal's triangle
without knowing the two elements directly above. Pascal presented
this problem at the end of his treatise as we have seen. To compute
the rth element in the nth row he gave the "explicit formula":
(~)
=
n(n - 1)(n - 2) · · · (n - r + 1)
l·2·3 .. ·r
Pascal's treatise states the formula in words, not in our modern
notation. He gives a remarkable proof.
Now the explicit formula does not apply, as it stands, to the case
r = 0. Yet we lay down the rule that, if r = 0, it should be interpreted
as
(~)
=
one of the boundary conditions.
49
1,
The explicit formula does apply to the case r = n and yields
(~)
=
n(n - 1)(n - 2) · · · 2 · 1
1 · 2 · · · (n- 1)n
= 1
which is the correct result for the other boundary condition.
We therefore have to prove the formula only for 0 < r < n, that is,
in the interior of the triangle where we can use the recursion formula.
Pascal states: "Although this proposition [the explicit formula]
contains infmitely many cases I shall give for it a very short proof,
supposing two lemmas.
The first lemma asserts that the proposition holds for the first
base, which is obvious. [The explicit formula is valid for n = 1,
because, in this case, all possible values of r, r = 0 and r = 1, fall under
the preliminary remark.]
The second lemma asserts this: if the proposition happens to be
valid for any base [for any value n] it is necessarily valid for the next
base [for n + 1].
We see hence that the proposition holds necessarily for all values
of n. For it is valid for n = 1· by virtue of the first lemma; therefore, for
n = 2 by virtue of the second lemma; therefore, for n = 3 by virtue of
the same, and so on 'ad infinitum.'
And so nothing remains but to prove the second lemma."
We translate his proof into modem notation:
We assume the formula works for the nth base. i.e. we have:
(~) =
n(n - 1)(n- 2) · · · (n- r + 2)(n - r + 1)
1 · 2 · 3 · · · (r - 1)r
and
n(n - 1) (n - 2) · · · (n - r + 2)
1· 2 · 3 · · · (r - 1)
50
Using our recursion formula:
(n;1) =
=
=
n(n - 1)(n - 2) · . · (n- r + 2)
1 · 2 · 3 · · · (r - 1)
[n- ~ + 1 + 1]
n(n- 1)(n- 2) · . · (n- r + 2) n+1
·-r
1 · 2 · 3 · · · (r - 1)
(n + 1)n(n - 1)(n - 2) · · · (n - r + 2)
1· 2· 3 · · · r
which is the explicit formula we wanted for
( n +r
1)·
Again we see Pascal putting the process of mathematical
induction to use.
We may also write the explicit formula. using the factorial sign.
as:
(~)
=
n!
r!(n - r)!
where n! = n(n - 1)(n - 2) · · · (3)(2)(1) and 0! = 1 by definition. (The
symbol n!. called n factorial. was first introduced by Christian Kramp of
Strasbourg in 1808 as a convenience to the·printer. It did not gain
immediate acceptance. and the altemative symbols Ln and 7t(n) were
used until late in the nineteenth century. Some English textbooks
suggested the reading "n - admiration" for n!. since the exclamation
point ! is a note of admiration.)
51
,)
4th row:
4!
(6) = 0!4!
=
4!
(i) = 1!3!
1
4!
=
4!
4!0!
=
4
4!
(~) = 3!1! = 4
(~) = 2!2! = 6
(4)
4
=
1
Because(~) represents the combination of n things taken rat a
time, this symbol is often seen in set theory, for it counts subsets. For
examble, given a set of five elements, say {A, B, C, D, E}:
the number of subsets with no elements: 0
(~)
= 5
the number of subsets with one element: {A},
{B}. {C}, {D}. {E}
(~)
= 10
the number of subsets with two elements:
{A,B}, {A,C}, {A,D}, {A,E}, {B,C}, {B,D}, {B,E},
{C,D}, {C,E}, {D,E}
(~) =
10
t..~e
number of subsets with t..hree elements:
{A,B,C}, {A,B,D}, {A,B,E}, {A,C,D}, {A,C,E},
{A,D,E}, {B,C,D}, {B,C,E}, {B,D,E}, {C,D,E}
(~)
= 5
the number of subsets with four elements:
{A,B,C,D}, {A,B,C,E}, {A,B,D,E}, {A,C,D,E},
{B,C,D,E}
(~)
= 1
the number of subsets with five elements:
{A,B,C,D,E}
53
Note that
(g) + (~) + (~) + (~) + (~) + (~) = 1 + 5 + 10 + 10 + 5 + 1 = 32 = 2 5 '
which not only gives the sum of the fifth row of the triangle, but also
the total number of subsets, proper and improper, of a given set of five
elements, i. e. given a set of n elements, the set has 2n subsets.
In 1827 the symbol (~) was used to designate binomial
coefficients. The next section of this paper shows the relation
between the theory of combinations and the formation of the binomial
coefficients.
54
Chapter 7
BINOMIAL COEFFICIENTS AND THE PASCAL PYRAMID
This chapter shows the relationship between the numbers in
Pascal's triangle and the binomial coefficients. the second topic Pascal
discussed at the end of his treatise. A fascinating generalization of the
binomial coefficients is evidenced by the Pascal pyramid, a subject
Pascal did not pursue. but is nevertheless difficult to overlook.
55
Binomial Coefficients
The numbers in Pascal's triangle are also known as binomial
coefficients. The general form of a binomial is (a+ b) and we consider
nonnegative, integral powers of the binomial:
(a!+ b)O
(a+ b)1
(a + b)2
(a + b)3
(a + b)4
(a+ b)5
(a + b)6
=1
= 1a + 1b
= 1a2 + 2ab + 1b2
= 1a3 + 3a2b + 3ab2 + 1b3
= 1a4 + 4a3b + 6a2b2 + 4ab3 + 1b4
= 1a5 + 5a4b + 10a3b2 + 10a2b3 + 5ab4 + 1b5
= 1a6 + 6a5b + 15a4b2 + 20a3b3 + 15a2b4 + 6ab5 +
1b6
The terms in the above expansions, apart from their coefficients,
can be written by inspection. For each expansion of
(a +b)n, we note the following:
( 1 ) Each expansion has n + 1 terms.
(2) The total degree of each term is n, i.e. in each term the sum of
the power of "a" and the power of "b" is n.
( 3) The first term is always an and the power of "a" decreases by 1
in each successive teun. The last teuu contains aO.
(4)
The first term contains bO and the power of "b" increases by 1 in
each successive term. The last term is bn.
( 5) The coefficient of the first and last terms is one. The coefficient
of any other term can be found by multiplying the coefficient of the
preceding term by the exponent of "a" in the preceding term and then
dividing by the number of terms that precede it.
56
(a + b)4
1
4
2
3
5
= 1 a4 + 4 a3b + 6 a2b2 + 4 a 1b3 + 1b4
example:
J,
1·4
1
J,
4·3
2
J,
J,
6·2
4·1
3
4
We now look at the coefficients from another angle. Consider
the above expansion. The five terms, without their coefficients, are:
Since (a+ b)4 = (a+ b)(a + b)(a + b)(a +b), each term in the expansion
is obtained by selecting exactly one letter from each of the four factors.
To get the first term, a4, we select no "b"s and four "a"s from the four
factors. Because this can be done in
(6) or one way, a4 occurs only
once in the expansion, and its coefficient is one. To get the second
term, a3b, we select one "b" from one of the factors and three "a''s
from the remaining three factors. This can be done in
(i) or four
ways. Thus the term a3b occurs four times and has the coefficient
four. Similarly, the term a2b2 is
ab3
in(~) or four ways;
obtaine~ in (~)or six ways;
the term
and the term b4 in(!) or one way. The
complete expansion of (a+ b)4 is the "sum" of all these terms:
We see that the coefficients are the elements of the fourth row in
Pascal's triangle. This process is generalized into the result known as
the "binomial theorem":
57
If n is a nonnegative integer. then
(a + b)n = (g}n +
Proof:
(~}n-lb + (~}n-2b2 + ... + (~}n-rtr + ... +(~}n
Each term in the expansion of
n factors
(a + b)n = (a + b)(a + b) · · · (a + b)
is of degree n in the variables "a" and "b." Thus. if we
ignore the coefficients. each term has the general
pattern
an-rtr. where r
= 0.
1. 2 .... n and (n - r) + r
= n.
The term an-rbr is obtained by selecting "b" from r of the
factors and "a" from the remaining n- r factors. This
selection can be made
an-rtr occurs
in(~) ways. Hence.· the term
(~)times and its coefficient is (~)
Therefore. the complete general term is
and the expansion is that shown in the theorem.
In summation notation this expansion may be written as:
n
(a + b)n
= Ii~}n-rtr
r=O
58
We note that the rth term of the expansion is:
n bn-(r-1)br-1
( r-1r
or
example:
·( n bn-r+1br-1
r-1r
(a + b)7
= (b}7 + (i}6b +
+
(~}5b2 + (~}4b3 + ~}3b4 + (~}2b5
(~}b6 + (~}7
The coefficients make up the seventh row of Pascal's triangle.
Had we only wanted to find the fifth term, we would have simply
calculated
with n = 7 and r = 5 to obtain:
59
example:
(2x - 3y)4
= (6)zx)4 + (i)zx)3(- 3y) +
(~)2x)2(-3y)2
+
(~)2x)(-3y)3
+ (!)-3y)4
=
1(l6x4) + 4(8x3)(-3y) + 6(4x2)(gy2) + 4(2x)(-27y3) + 1(81y4)
=
16x4 + -96x3y + 216x2y2 + -216xy3 + 81y4
Had we only wanted to find the third term we would have simply
calculated
with n
=4
and r
=3
to obtain:
( 3: 1 )zx)4-3+ 1 (-3y)3-1
= (~)2x)2(-3y)2
= 6(4x2)(gy2) = 216x2y2
Since the coefficients of the expansion of (a + b)n are the
elements in the nth row of Pascal's triangle, which are
(~) is known as the binomial coefficient symbol. Euler designated the
binomial coefficients by
rr) in a paper written in 1778, but not
published until 1806, and used the
device[~] in a
60
paper of 1781
published in 1784. The
notation(~} which was to become the more
common convention, appeared about 1827.
In 1664 Newton discovered the general binomial theorem, or
expansion of (a+ b)n, where n may be a fractional or a negative
exponent as well as a nonnegative, integral exponent. The resulting
expansions are infinite series.
It is interesting to note that we can show directly that the
coefficients of the expansion of (a + b)n are the numbers in Pascal's
triangle. We take a different approach using the principle of
mathematical induction.
We look at the first few expansions.
(a+ b)O
=1
(a+b)1 =a+b
(a+ b)2 =
a+ b
x a+b
ab+b2
a2 + ab
a2 + 2ab + b2
(a + b)3 = (a + b)2(a + b) =
a2 + 2ab + b2
x
a + b
a2b + 2ab2 + b3
a3 + 2a2b + ab2
(a + b)4 = (a + b)3(a + b) =
a3 + 3a2b + 3ab2 + b3
x
a
+ b
a3b +3a2b2 +3ab3 + b4
a4 + 3a3b +3a2b2 + ab3
61
The expansions of (a+ b)n for n
= 0,
1, 2, 3, 4 give the coefficients:
1
1
3
1
4
1
2
1
1
1
3
6
1
4
1
We would like to show that this process continues, i.e., the coefficients
are the numbers in Pascal's triangle and that the coefficients of the
expansion of (a + b)n+ 1 can be obtained from the coefficients of
(a + b)n using the same rule to find the numbers in Pascal's triangle,
i.e. Pascal's rule.
Proof:
We have already seen that for n = 0, 1, 2, 3, 4 the
coefficients are the numbers that make up the first five
rows of Pascal's triangle. We also note that the coefficients
of an and bn are always one.
Consider the expansion of (a+ b)n.
Let (a+ b)ll =
where c 1 , c 2 , c 3 , ... , cn_ 1 are the coefficients of the
terms an-~r for r = 1, 2, 3, ... n-1
Now, (a + b)n+ 1
= (a + b)n(a + b) =
an+ c1 an-1 b + c2an-2b2 + c3an-3b3 + ... +
cn-1 abn-1+ bn
+ b
anb + c 1an-1b2 + c2 an-2b3 + c3 an-3b4 + ... + cn-1abn + bn+1
x
a
an+1 + c1 anb + c2a_ll-1b2 + c3an-2b3 + c4an-3b4 + ... +
abn
an+1 + (1 + c1 )anb + (c1 + c2)an-1b2 + (c2 + c3)an-2b3 + (c3 + c4)an-3b4 +...
+ (c
n-1
+ 1)abn + bn+1
62
Q
Now, the coefficients for (a + b)n are:
1
1
and the coefficients for (a+ b)n+1 are:
i.e. the coefficients of the expansion of (a+ b)n+1 are obtained from
the coefficients of the expansion of (a + b)n using Pascal's rule.
Although Pas~al discussed using his arithmetic triangle for
taking powers of binomials using only nonnegative integers for powers,
we will investigate the negative, integral powers since the triangle may
still be used to find the coefficients of the terms of the expansions.
We obtain the following expansions by "brute-force" long division:
63
•
't
We now look at Pascal's triangle in
columns lined up:
inv~rted
form with the left-hand
1
8
28
56
70
56
28
8
1
7
21
35
35
21
7
1
1
6
15
20
15
6
1
1
5
10
10
5
1
1
4
6
4
1
1
3
3
1
1
2
1
1
1
1
c
II
a
+ b'
1
Pascal's rule translates to the L-shaped rule to generate a row
from the row underneath (since we are upside down), that is, "a+ b =
c". But it's just as true that b = c - a. This rule can be used to fill out
each row with zeros to the right. We can also expand it downward by
the same rule. ~~~~~~-!l:t~13~--~~:~.13.!~I?:I3.~~ ()P!~~---~~- !t?!J9.~!!g
chart and each new row can have as many numbers added as we wish,
uf;a-repeati~g-cieciffiaE
_____ -·----·· --
·- ·
~---------~-~---'-....-c~.,...---•-~~·.,_.,••k-•«~~.~,.._,.,._~--.,._...,or.,,~-',.,_,_,..,._,,,,
64
r
I(~)
8
7
6
5
4
3
2
1
n
0
-1
-2
-3
-4
-5
-6
-7
-8
0
1
2
3
4
5
6
7
8
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
8
7
6
5
4
3
2
28
21
15
10
6
3
1
0
0
1
3
6
10
15
21
28
36
56
35
20
10
4
1
0
0
0
-1
-4
-10
-20
-35
-56
-84
--120
70
35
15
5
1
0
0
0
0
1
5
15
35
70
126
210
330
56
21
6
28
7
1
0
0
0
0
0
0
1
7
28
84
210
462
924
1716
8
1
0
0
0
0
0
0
0
0
1
0
-1
-2
-3
-4
-5
-6
-7
-8
1
0
0
0
0
0
-1
-6
-21
-56
-126
-252
-462
-792
1
0
0
0
0
0
0
0
-1
-8
-36
-120
-330
-792
-1716
-3432
1
9
45
165
495
1287
3003
6435
[Source: 4, p. 449]
If we name each row by an integer, n, and each column by a
nonnegative integer, r, we may denote each entry in the array by the
symbol(~} We have already seen that this symbol generates the
coefficients of the expansion of (a + b)n where n is an integer, n
~
0,
and 0::; r::; n. Pascal defined (~)to equal
--
>'.
)
\
n(n - 1)(n - 2) · · · (n - r + 1)
\!
1· 2 · 3 · · · r
and we have shown that this is equivalent
to(~)= r!(nn~ r)!
where n! =
n(n - 1)(n - 2) · · · 2 · 1.
Since n! has no meaning for numbers other than nonnegative
integers, we use,:-Pascal's definition for negative integers.
\
.
examp 1es:
(a l
(-34)
=
(-4)(-5)(-6)
. .
1 2 3
=
-20\\
,PI·
65
(b)
-5) = (-5)(-6)(-7)(-8) =
70
( 4
1·2·3·4
{
l
.
''"•· "
-
-·
.. ..
-~--
When n < 0, the sign
when r is odd,
-
.-
'.
__,_,/)
of(~) may be positive or negative. Since
(~)is negative, and when r is even, (~)is positive,
(-l)r determines the sign of(~). Also, for any column, r, the absolute
value of every entry in the lower portion of our chart occurs in the
upper portion. Hence we need only establish the row in which the
absolute value occurs. For example, if n = -4 and r = 3,
(~) = -20 and
l-20
I = 20 appears in row n
= 6. It is clear that when
n < 0, the absolute value of the entry in the rth row appears in the
(r- n -1)th row above. Thus we have a second way to define(~)
lr')
=
(-1)r(r-~- 1 J when n
e J, n < 0,
r~
0 and we may now express
the binomial theorem for n e J, n < 0, as:
00
(a + b)n =
L(~}n-~r
r=O
(as opposed to (a+ b)n =
~lr}n-Jbr
for n e J. n;, 0)
r=O
Note, however, that we may also use this new summation for
n e J, n ~ 0, since(~)= 0 when r > n. The infinite series becomes a
finite one since after a certain point all the terms are equal to zero.
When n< 0, (a+ b)n represents an infinite series, and there is a
problem of convergence since the terms do not in general become
identically zero after some point.
66
We investigate the possibility of convergence for this infinite
series. We have
We may also write the terms as follows:
t1
(a+ b)n =
t2
t3
t4
an + nan-1b + n(~~ 1 ) an-2b2 + n(n-;~(n- 2 ) an-3b3
ts
+
n(n-1)(n-2)(n-3) n- 4 b 4
4!
a
+ ...
Consider the ratio of each pair of successive terms:
b
= n.a
n(n-1) n-2b2
2! a
n-1 b
=
2'a
n(n-1)(n-2) n- 3 b 3
3!
a
n(n-1) n-2b2
2! a
n-2 b
= --s·a-
n(n-1)(n-2)(n-3) n- 4 b 4
4!
n-3 b
n(n-1)(n-2) n- 3 b 3
= 4 ·a
3!
a
tm+1
and in general, tm
b
= n-(m-1)
.m
a
67
We now look at the limit of the sequence of these ratios:
lim [n-(m-1). ~l
m-+oo
m
aj
b
b
= (0- 1 + 0) . -a = - a-
lim ltm + 11
By the ratio test, if m-+oo
ltml
< 1,
then the series t 1 + t2 + t3 + ... + tm + tm+ 1 + ... converges absolutely.
Since
lim ltm + 1l
1- bl
=
lal
m-+oo ltml
lbl
=TaT·
our series will converge absolutely when
lbl
laT < 1 or I b I
< Ia I .
i.e. for n e J-, the infinite series, (a+ b)n, converges absolutely when
lbl < lal.
00
Note: if "a" and "b" are switched, then (b + a)n =
Ll~pn-rar
r=O
converges when I a I < I b I .
If we have our chart available, the coefficients of the expansion of
(a + b)-3 are the entries in row n = -3; the coefficients of the
expansion of (a + b)-4 are the entries in row n = -4, etc. However, the
rows in our chart for n e J, n < 0, are the same as the diagonals in
68
Pascal's triangle except for the negative signs, so we may return to our
simpler array which we used for n e J, n ;::: 0:
rO
rl
r2
To summarize for all integral values of n: the coefficients of the
terms in the expansion (a + b)n for n ;::: 0 are the entries in the nth
row (we have a finite number of entries in each row to give us a finite
expansion.) The coefficients of the terms in the expansion (a + b)n for
n < 0 are the entries in the I n + 1 Ist diagonal provided we alternate
the signs beginning with positive (we have an infinite number of
entries in each diagonal to give us an infinite expansion.)
The following diagram summarizes our findings:
..........................................
1111111111111111111111111111 . . • • • • • • • •
1···
1
row
row
row
row
row
row
row
row
row
0 : (a +b) 0
1 : (a + b) 1
2 : (a +b) 2
3 : (a + b) 3
4 : (a + b) ~
5: (a+ b)
6 : (a + b) ~
7: (a+ b)
8 : (a + b) 8
.
..
row n : (a+ b) n
....0..
~
....0..
~
....
0..
p_.
&
&
....
-.J
0)
en
~
..........
...-..
PI
...-..
PI
+
+
+
...-..
PI
+
...........
...........
...........
...........
tr'
I
co
tr'
I
-.J
PI
tr'
I
0)
tr'
I
en
0..
.....
....0..
w
.... ....0..
0..
~·
&i
1\)
1-'
0
...-..
.........
.........
.........
+
+
+
+
...........
...........
...........
...........
I
I
&i' &i
PI
tr'
.
PI
0"'
I
w
~
69
PI
0"'
1\)
PI
0"'
I
1-'
The Pascal Pyramid
There seems to be no evidence of Pascal having developed a
pyramid to correspond to his triangle. It is rather interesting to note,
however, that there is a pyramidal pattem for determining the
coefficients of a trinomial raised to successive powers, just as Pascal's
triangle is a pattem for determining the coefficients of the binomial
expansion.
We consider the following:
(a+ b + c)O
=1
(a+b+c)1
=1a+ 1b+ 1c
(a + b + c)2 = 1a2 + (2ab + 2ac) + (lb2 + 2bc + 1c2)
(a + b + c)3
= 1a3
+ (3a2b + 3a2c) + (3ab2 + 6abc + 3ac2)
+ (1b3 + 3b2c + 3bc2 + 1c3)
(a + b + c)4 = 1a4 + (4a3b + 4a3c) + (6a2b2 + 12a2bc + 6a2c2)
+ (4ab3 + 12ab2c + 12abc2 + 4ac3)
+ (1b4 + 4b3c + 6b2c2 + 4bc3 + 1c4)
(a + b + c)5 = 1a5 + (5a4b + 5a4c) + (10a3b2 + 20a3bc + 10a3c2)
+ (10a2b3 + 30a2b2c + 30a2bc2 + 10a2c3)
+ (5ab4 + 20ab3c + 30ab2c2 + 20abc3 + 5ac4)
+ (1b5 + 5b4c + 10b3c2 + 10b2c3 + 5bc4 + 1c5)
The first layer of the pyramid, the vertex, comes from
(a + b + c)O = 1. We look at the following layers both in terms of
variables and in terms of coefficients.
70
1
n
= 1:
1
n
=2:
1
n
1
2
= 3:
c.
3
1
1
1
71
n = 4:
4
b
,c..__ _ _ __ , ._ _ _ _ _.....__ _ _ ____,...___ _ _ _->o.
b 2 c2
b3 c
c4
bc3
1
1
4
6
72
4
1
n =
5:
1
1
5
10
10
73
5
1
Upon inspection, the variables and their powers fall into a
definite pattern. The exponents of the variables can be determined by
assigning the highest power of "a" to the first row. The product of the
second highest power of "a" times the first power of each of the other
variables is assigned to the second row. In the next rows, we continue
to decrease the power of "a" and arrange the powers of the other
variables so that the sum of the powers equals n. For each row the
power of "a" will remain the same, the power of "b" will decrease, and
the power of "c" will increase.
We now investigate the coefficients. The following is the
pyramid created by the first six levels. We then look at each level with
respect to the level above it to get our "rule" for finding the
coefficients from the level above.
74
1
5
10
10
5
The First Six Levels of Pascal's Pyramid
75
1
n = 1:
1
= 2:
n
1
1
2
1
n = 3:
1
3
1
3
1
= 4:
n
1
s
4
76
4
1
1
n = 5:
1
5
10
10
5
1
The method for constructing the coefficients is as follows: the
circled numerals at each intersection are the coefficients for the given
n. To get them, insert the numerals for the coefficients for n - 1, that
is the numerals from the previous layer of the pyramid. These symbols
are placed in the triangles that "point up." At each intersection inside
the large triangle, there are three of these small triangles; at each
intersection on the sides of the large triangle, there are two; at each
vertex of the large triangle, there is one. In each case, the sum of the
numbers in the small triangles that meet at any intersection is the
coefficient for that intersection.
To prove this recursive property of the pyramid requires first
that we are able to state the property in formal, mathematical
language. This requires a suitable symbol for the trinomial
coefficients.
We saw for the binomial expansion of (a+ b)n that(~) gives the
number of ways in which a set of n elements can be partitioned into an
ordered pair of subsets having r and n - r elements, respectively. That
is, since there are
(~)ways to select "b" from r of the factors and "a"
from the remaining n - r factors, the coefficient for each term of the
77
expansion, an-Ibr, is(~} We seek similar notation for a more general
problem, to fmd the number of ways in which a set of n elements can
be partitioned into an ordered set of k subsets having r1, r2 ... , and rk
elements, respectively with r1 + r2 + ... + rk = n. Since the first
subset can be selected from the entire set in
(r~) ways,
the second
subset can subsequently be selected from the remaining n- r1
elements in lrn-rl)
r
ways, the third subset can be selected from the
2
remaining n - r1 - r2 elements in
lrn- rlr 3 - r2) ways, and so on, the
desired number of partitions is
n!
(n-r 1)!
(n-r1-r2)!
(rk)!
= (n-r1)!r1! · (n-r1-r2)!r2! · (n-r1-r2-r3)!r3! · · · (rk-rk)!rk!
Symbolically, we write this number of partitions as
and call it a multinomial coefficient as it is the coefficient of
x 1r1x2r2 · · · xkrk in the expansion of (x 1 + x2 + ... + xk)n.
For the expansion of our trinomial, (a+ b + c)n, we let
n!
(i + j + k)!
n )
. .
i . k
.
., ., k' =
., J"' k'
.
(I. . J. • k equal the coefficient of a hJc which equals I.J..
1. • •
Now, we can also let (i,
f k) denote the point that is i units from
the base of one of our triangles joining en to an, that is j units from the
78
base joining an to bn, and that is k units from the base joining bn to en
(the "unit" being the altitude of one of the small triangles).
As an illustration, we label the coordinates of the vertices for the
triangle n = 4 using the (i,
f k) or simply (i, j, k) notation.
Recall that the recursive formula, Pascal's rule, for constructing
the binomial coefficients as well as the elements of Pascal's triangle is
1
1
given by (n; ) = (r~1) +(~)or(~)=(~~~)+ (n; } Similarly, we show
that an analogous construction of the trinomial coefficients can be
made using the recursive rule
n ) ( n-1 ) ( n-1 ) ( n-1 ')
( i,j,k = i-1,j,k + i,j-1,k + i,j,k-1)
where i + j + k = n. This will be interpreted as the point (i, j, k) on
· the nth level is centered under the triangle in the (n - 1)st level
whose vertices are (i - 1, j, k), (i, j - 1, k), and (i, j, k - 1) and the sum
of the trinomial coefficients associated with the vertices of the upper
triangle is equal to the trinomial coefficient found directly below that
triangle.
n-1 ) ( n-1 ) ( n-1 )
( i-1,j,k + i,j-1,k + i,j,k-1
Proof:
=
i+j+k-1) (i+j+k-1) (i+j+k-1)
"1"k+
""1k+
""k1
( 1'J,
1,
J- •
1,
J, -
79
=
(i + j + k - I)!
(i + j + k - 1)! (i + j + k - I)!
(i - I)! j! k! + i! 0 - I)! k! + i! j! (k - I)!
=
i(i + j + k - I)!
j(i + j + k - I)!
k(i + j + k - I)!
i! j! k!
+
i! j! k!
+
i! j! k!
=
(i + j + k)(i + J + k - I)!
., ., k!
1. J.
=
(i + j + k)! _ (i+j+k) _ ( n )
i! j! k!
- i, j, k - i, j, k .
This recursive rule works for boundary points as well as for
interior points if we extend our coordinate system in a natural way by
introducing negative coordinates as shown in the following diagram
and by setting (i,
J. k) = 0 whenever the point (i, j, k) lies outside the
main triangle; that is, when any coordinate is negative.
..
(-1,0,5)
,, ' ' '
,
.
'
n = 4:
::_(9~~~~)____\~. (-1,1,4)
.. .
,,
, ''
''
'
~~:------..-'J~~1-'~L----'..... (-1,2,3)
,, ' ' '
,
,
'
2
lk-...:;____;;,....,..~E-------*·:J~ ~ .!~L--- ~ ~
..
,
.. ,
(- L3,2)
''
''
1
. _ ._ _;.....__,.f--;....._--'"""'"*--"""'----4~ ~ _(9~~~_ _)____ ~::.,
,
.
, ..
,~ (0,4,0)
.----~~---.........---'--"""'----¥----'----'-~------
80
(-1 A,l)
''
''
',
-------.
n
= 5:
We reproduce the two triangles on pages 76 and 77 for n = 4 and
n = 5 so we can use these in combination with our last page to check
some results.
1
1
4
6
81
4
1
1
n =
5:
5
1
10
10
5
1
1) Check an interior point for n = 5, say (2, 1, 2):
a) (2, 1, 2) has the coefficient 30 which is the sum of the three
vertices of the triangle above it in level 4. (i. e. 30 = 12 + 12 + 6)
.
4!
4!
4!
Since 1! 1! 2! + 2! 0! 2! + 2! 1! 1!
= 12 + 6 + 12 = 30
5!
=
2! 1! 2!
b) The coordinate (2. 1, 2) is centered under the triangle in level
4 whose vertices are (1, 1, 2), (2, 0, 2), and (2, 1, 1).
2) Check a boundary point for n = 5, say (0, 2, 3).
a) (0, 2, 3) has the coefficient 10 which is the sum of the two
vertices of the triangle above it in level 4 (or three vertices if we
include the vertex with the negative coefficient).
(i. e. 10 = 4 + 6 + 0)
82
or ( o.
~. 3) = (-1, ~. 3) +
( 0,
i. 3) + (0, ~. 2)
4!
4'
since 0 + 0! 1! 3! + 0! 2i 2!
=0
= 10
+4 +6
5!
= 0! 2! 3!
b) The coordinate (0, 2, 3) is centered under the triangle in level
4 whose vertices are (-1. 2, 3). (0, 1, 3). and (0, 2, 2).
The three-dimensional nature of the Pascal pyramid makes it
awkward to use for calculating the trinomial coefficients. However, we
do arrive at a neat way of determining those coefficients directly from
the Pascal triangle. We look at level 4:
1
1
4
6
83
4
1
,, .
We note that the "lines" of the array are "reducible to lowest
terms," and on carrying out such a reduction we arrive at the first five
lines of Pascal's triangle:
1
1
4
6
4
1
It is easy to see why this is so; it follows by treating the trinomial
expansion as a binomial expansion:
(a + b + c)4
=
= ((a + b)
+ c)4
(6) (a+ b)4 + (i) (a+ b)3c +(~)(a+ b)2c2
+(~)(a+ b)c3 + (!) c4
=
l(a + b)4 + 4c(a + b)3 + 6c2(a + b)2 + 4c3(a + b) + Ic4
Therefore, to obtain the coefficients for the expansion of
(a + b +c)n we have the following rule: Construct the Pascal triangle
down to the nth level. Multiply the rows successively by the numbers
found in the bottom line. That is, multiply the first row by the first
element in the bottom row, the second row by the second element in
the bottom row, and so on.
For example, the coefficients for the terms of the expansion of
84
(a + b + c)5 are obtained by writing out the first six rows (the Oth to
the fifth row) of Pascal's triangle and performing the multiplication
described above:
1
1
1
1
1
1
3
1
4
6
10
5
1
2
4
1
1
3
1
X5
X 10
X 10
X5
X 1
X
10
1
5
1
=
5
10
10
1
5
10
20
30
20
5
5
20
30
10
10
30
10
5
1
5
Evidently one needs four-space to determine an array for the
coefficients of the expansion of (a+ b + c + d)n. However, following
the multiplication device and using the nth row of the triangle times
the numbers in each layer of the pyramid, these coefficients can be
determined. The array, however, will not be symmetrical. (Maybe it is
in four-space.) The procedure for determining the variables is the
same as in the pyramid.
example: (a + b + c + d)2
= a2 + (2ab + 2ac + 2ad) + (b2 +2bc + 2bd + c2 + 2cd + d2)
85
The array is formed as follows:
Layers of the pyramid
ton= 2:
Second row
of the triangle:
Coefficients:
1
X
1
=
1 1 1
1 2 2 1 2 1
X
2
X
1
= 2 2 2
= 12 2 12 1
1
example: (a + b + c + d)3
= a3 + (3a2b + 3a2c + 3a2d)
+ (3ab2 + 6abc + 6abd + 3ac2 + 6acd + 3ad2)
+(b3 + 3b2c + 3b2d+ 3bc2+6bcd+ 3bd2+ c3+ 3c2d+ 3cd2 + d3)
The array is formed as follows:
Layers of the pyramid
ton= 3:
Third row
of the triangle:
1
1 1 1
1 2 2 1 2 1
1333631331
X
X
X
X
1
3
3
1
Coefficients:
=
1
=
3 3 3
=366363
=1 3 3 3 6 3 1 3 3 1
The process continues to apply to raising the sum of k numbers
to successive powers.
86
Chapter 8
PROBABILITY AND THE PROBLEM OF POINTS
In chapter 6 we saw Pascal's work with combinations. In
modern texts one rarely sees a discussion of combinations without a
discussion of probability. Just as the numbers in Pascal's triangle can
be linked with combinatorial theory, it is also linked with probability
theory. This chapter examines Pascal's third topic, the problem of
points.
87
Probability
Historically, the theory of probability was initiated as an aid to
gamblers. The present widespread interest in the subject, however, is
not evidence of tremendous gambling activity. Rather, the permeation
of statistical methods into problems of industry, economics, insurance,
medicine, sociology, and psychology raised questions that had never
arisen in previous applications of mathematics and that could be
answered only by a theory of probability.
The symbol (~)is often seen in problems of probability. We shall
look at examples involving coin flipping (or any other event with two
equally possible outcomes.)
88
#coins
flipped
Possible
outcomes
#heads in
each
outcome
#ways to get
the following
#of heads:
#of total
outcomes
01234
0
1
2
3
4
T
H
TT
TH
HT
HH
TTT
TTH
THT
HTT
THH
HTH
HHT
HHH
TTTT
TTTH
TTHT
THTT
HTTT
TTHH
THTH
THHT
HTTH
HTHT
HHTT
THHH
HTHH
HHTH
HHHT
HHHH
0
1
1
0
1
1 1
2
= 20
= 21
0
1
1
2
1 2 1
4
= 22
0
1
1
1
2
2
2
3
1 3 3 1
8
= 23
0
1
1
1
1
2
2
2
2
2
2
3
3
3
3
4
1 4 6 4 1
16
= 24
89
From the fourth column we obtain the entries in Pascal's triangle
and the number of possible outcomes in the fifth column is equal to
the sum of each row in the triangle (2n):
1
1
2
1
3
1
1
1
4
1
3
6
1
4
1
(g)
=
(~)
(6)
(~)
(i)
(~)
(~)
(i) (~) (~)
(6) (i)
(~) (~)
(!)
We now look at the "probability" of coming up with a certain
number of heads for each case.
90
@
#coins
#of total
outcomes
Probability of coming
up with the following
#of heads:
1
0
2
3
0
1
1
1
2
2
1
2
2
4
1
4
2
4
1
4
3
8
1
8
3
8
3
8
1
8
4
16
4
16
6
16
4
16
Sum of
probabilities
4
1
1
1
T6
1
1
1
1
16
1
Our probability .triangle below is obtained from Pascal's triangle
by dividing each entry by the total number of possible outcomes for
each case. (The sum of the elements in each row of the triangle.)
Arithmetic triangle
Row sums
Probability triangle
1
1
1
1
1
1
2
3
4
1
1
3
6
2
= 20
= 21
4
= 22
1
4
1
8
= 23
16
= 24
1
1
2
1
4
1
8
1
16
1
2
2
4
3
8
4
16
1
4
1
8
3
8
6
16
4
16
1
16
Note that the sum of each row in the probability triangle is one,
which agrees with the fact that the sum of the probabilities of all the
possible outcomes of any event is always one.
91
•
Each entry of the arithmetic triangle is the sum of the two
entries above it. Each entry of the probability triangle is the "average"
of the two entries above it.
To prove this relationship easily, we take any three entries, a, b,
c, from the probability triangle such that "a" and "b" are any two
adjacent row elements, and "c" is the element in the next row
between "a" and "b":
.... a b ....
. . . . . . . . . c ........ .
These elements have respective counterparts from the arithmetic
triangle; call them A, B, and C. Now we know that by Pascal's rule
A + B = C. Also,
A
B
a = - andb=2n
2n
where n is the number of the row in which "a" and "b" lie (n is also
the number of the row in which A and B lie.) Since "c" is in the
(n + 1)st row, we have
c
c = 2n+1
Now, since A+ B = C, we have
2n . a + 2n · b = 2n + 1 · c
2n . a 2n . b
2n + 1 . c
2n+1 + 2n+1 = 2n+1
(a+ b)
2
= c
i.e. c is the "average" of "a" and "b", which is what we wanted to show.
92
The Problem of Points
Pascal's interest in the science of probability arose from a
problem that may not have seemed significant at the time, but was all
too human: how does one divide the stakes in an interrupted game of
chance if the players are of equal skill? The "Problem of the Points"
had been addressed by Pacioli in 1494, by Cardan in 1539, and by
Tartaglia in 1556. The problem remained in most arithmetic texts
well into the 1600s. A real advance was not made, however, until
1654 when the problem was proposed to Pascal by the Chevalier de
Mere, a gambler of unusual ability who confessed that he could not
solve the problem in the general case. Pascal communicated the
problem to Fermat. There ensued a remarkable correspondence
between the two men, in which the problem was correctly, but
differently, solved by each. The correspondence was fundamental to
the development of modern concepts of probability. Pascal looked
upon the problem as a problem in arrangements or combinations of
given things and in counting these arrangements. With characteristic
insight he selected the arithmetical triangle for handling the subject.
From a modem viewpoint, it may seem ironic that a man of Pascal's
spiritual gifts would concern himself with a problem related to
gambling, but his interest was probably philosophical and
metaphysical. [41, p. 89]
We restate the "Problem of Points": two players each need a
certain number of points in order to win; if they separate without
playing out the game, how should the stakes be divided between
them? The qu~stion amounts to asking what is the probability which
each player has, at any given stage of the game, of winning the game?
The first observation Pascal makes is that "two players who are
playing for a set of two games with a score of 1 to 0 are in the same
position as two others who are playing for a set of three games with a
score of 2 to 1. For both couples have this in common that in order to
win, one player needs only one game and the other, two. And therein
consists the odds, which should determine divisions; so that strictly
93
only the number of games left for each to win should be considered
and not the number already won." [23, p. 461]
For simplicity, we initially consider the proper division of the
stake in an interrupted game in which A and B are playing and four
points are needed to win. Let M be the stakes.
First, we shall discuss five specific cases and then proceed to a
solution in the general case.
Case 1. A needs no points to win. That is A has already won.
Clearly, A is entitled to all of M. (Note: B may need 1, 2, 3, or 4
points to win.)
Case 2. B needs no points to win. That is B has already won.
Clearly, A is entitled to none of M. (Note: A may need 1, 2, 3, or 4
points to win.)
Case 3. A and B each need one point to win. Clearly, M should
be equally divided, making A's fair
share~·
Case 4. A needs one point and B needs two points to win. The
next move has two possible outcomes:
(a) A wins the point and is entitled to all of M.
fo) B wins the point and we are back to case 3 where each
needs one point to win and A's fair share is
~. Since A can be
M
expected to win M half the time and 2 half the time, we say A's
"expected share" is the average of the two. Thus, A's fair share is
M) = 3
21 ( M+2
4 M.
Case 5. A needs two points and B needs one point to win. This
is the reverse of case 4. So A's expected share is
94
We can easily develop a recursive formula for A's fair share,
SA(a, b), where A needs "a" points to win and B needs "b" points to
win. The entries below are of the form (a, b, SA(a, b)). The underlined
entries are the outcomes covered in our five cases.
[Source: 9, p. 377]
We note that the third coordinate in each entry is the average of
the two third coordinates in the two entries above. For example, from
case 4 we have
(1. 2, ~)and
~ = ~ (M+~)
Our recursive formula for our example is
95
In general, the recursive formula is
SA(a, b)
= 21
(SA (a-1, b)+ SA (a, b-1)).
Pascal used his arithmetic triangle to develop a nonrecursive
formula. Consider the case SA (2, 3) in which A needs two points and
B needs three points to win. From the diagram, we see that
SA (2, 3) =
~ ~ M.
In this case the game would be over in at most four
more points (for A orB to win). Now a four-point game has 24
possible outcomes:
BBBB
ABBB
BABB
BBAB
BBBA
.MBB
ABAB
AAAB
.MBA
ABBA
ABAA
BMB
BABA
BAAA
AAAA
BBAA
1 way for
A to win
0 points
J_
(6)
4 ways for 6 ways for
A to win
A to win
1 point
2 points
4 ways for
A to win
3 points
J_
J_
J_
(i)
(~)
(~)
A can win two or more points in
1 way for
A to win
4 points
J_
(!)
(~) + (~) + (!)ways.
These
results represent theoretical possibilities even though it may not be
96
necessary to play all four points. Consequently, the probability that A
will win the game is given by
1 [
] 11
=16 6 + 4 + 1 =16
and A's fair share
is~~
M which agrees with our recursive formula for
SA (2, 3).
We now generalize this for a contest of n points with stake M in
which A needs "a" points to win and B needs "b = n - a + 1" points to
win. A's fair share of the stake Misgiven by
S(n) = M if a = 0}
S(n) = 0 if b = 0 trivial cases .
Proof:
Even though we use modem notation, the method of
mathematical induction follows that of Pascal's.
For the non-trivial case: For n = 1 we have a= 1 and b = 1
(each needs one point to win).
S(1) =
~ [(~ )] = ~
· 1=~
which is correct since if each of A and B needs only one
point to win then the stake should be divided equally.
Assume the statement S(n) is true for n = N. That is,
97
assume that when a + b - 1 = N, S(N) gives A's fair share of
the stake. That is
We must show that the statement is true for
a+b-l=N+l.
We have two cases and we use the recursive formula,
1
SA(a, b) = 2 (SA (a-1, b)+ SA (a, b-1)].
for both.
easel: b = 1
Since a + b - 1 = N + 1
a +1 - 1 = N + 1
a- 1
=N
1
Thus, SA(a, b)= 2 (SA (a-1, 1) +SA (a, 1-1))
1
= 2 (SA (N, 1) + SA (a, 0)]
= 21 SA
(N, 1)
1 M[ (N)~
. . assumption
.
N U by Induction
=2 · N
2
=S
(N + 1)
98
i1
Case 2: b > 1
1
SA(a, b) = 2 [SA (a-1, b)+ SA (a, b-1))
Using Pascal's rule for his triangle,
(r~1) + (~) = (n;
1
)
(i.e. each entry is the sum of the two entries above), and
the fact that
SA(a,b)=
2
(~) = 1 = (~: ~ ). we have:
M
[(N +
+ 1) + (N
+ 1)~
N+1
a 1) +···+ (NN
N+1U
= S(N+l)
Our proof is complete by use of mathematical induction.
99
•
Chapter 9
PATTERNS
This chapter, in addition to dealing with Pascal's last topic,
order of numbers, is devoted to citing a few of the patterns found in
the array of numbers we call Pascal's triangle. Pascal, himself, noted
some of these patterns, but it is no surprise to find that
mathematicians after him found many additional patterns embedded
in this array. We divide these seemingly countless patterns into seven
categories: rows, columns, and diagonals; hexagons; Fibonacci
sequences; primes; multiples; figurate numbers; and higher
dimensional figurate numbers.
While some of the patterns are trivial and quite obvious, they are
nonetheless curious, especially for those meeting the triangle for the
first time. Some of the patterns have shown up in earlier chapters and
are included for completeness.
100
Rows. Columns. and Diagonals
The following sixteen patterns are only a sampling of the many
patterns found in the rows, columns, and diagonals of Pascal's triangle.
As we saw in chapter 5, c 0 = d 0 , c 1 = d 1, and so on. This is
illustrated below.
do co
1 c1
d2 1 1 ~
d:3 1 2 1 c 3
d4 1 3 3 1 c4
ct 5 1 4 6 4 1 c5
d 6 1 5 lO 10 5 1 c6
d1
'7
1 6 15 20 15 6 1 ~
d8 1 7 21 35 35 21 7 1 C8
1 8 28 56 70 56 28 8 1
Therefore, any pattern that holds for columns will also hold for
diagonals.
101
1. The sum of the elements in the nth row is 2n. (This is
Consequence
8... in Pascal's treatise.)
-·---··
.... .
____
------------·~--·---~
--
-_..,
...
-
-/
Row Sum
Row
1
0
1
1
1
2
1
4
1
5
6
7
8
1
1
8
1
6
7
21
28
3
6
15
35
56
4
20
35
70
= 16 = 24
1
5
15
21
56
= 2 = 21
= 8 = 23
1
10
10
= 1 = 2°
= 4 = 22
1
2
4
5
I
1
3
1
3
I
= 32 = 25
1
6
= 64 = 26
1
7
28
=128=27
1
8
1
= 256 = 28
n
2. The sum of the elements in the nth row is double the sum of the
elements in the preceding row. (Consequence 7)
The first pattem shows this. i.e. the sum of the elements in row n is
2n = 2 · 2n-l, which is twice the sum of the elements in the (n-l)st
row.
102
3. The sum of all the elements of row 0 through row n, inclusive, is
2n+1 - 1. (Consequence 9)
0
1
2
3
4
1
1+(1+1)
3 + (1 + 2 + 1)
7 + (l + 3 + 3 + 1)
15 + (1 + 4 + 6 + 4 + 1)
= 1 = 20+ 1 - 1 = 2 - 1
=3=21+1-1=4-1
= 7 = 22+ 1 - 1 = 8 - 1
= 15 = 23+1 - 1 = 16- 1
= 31 = 24+1 - 1 = 32 - 1
= 2n+1 - 1
n
4. In any row, except ro. the result obtained by alternately subtracting
and adding is always zero.
1
1
1 1 - 7
1 - 8 +
1 - 1
=0
1~2+1
=0
=0
=0
= 0
= 0
= 0
= 0
1- 3+3- 1
1-4+6-4+1
- 5 t 10- 10 + 5 - 1
6 + 15 - 20 + 15 - 6 + 1
+ 21 - 35 + 35 - 21 + 7 - 1
28 - 56 + 70 - 56 -t 28 - 8 + 1
103
5. In any row, except ro. the result obtained by altemately adding and
subtracting is always two.
1
1 + 1
1+2-1
1-+3-3+1
1-+4-6+4-1
1 + 5 - 10 t 10 - 5 + 1
1 + 6- 15 -t 20- 15 + 6 - 1
1 + 7 - 21 + 35 - 35 + 21 - 7 + 1
1 + 8 - 28 + 56 - 70 t 56 - 28 + 8 - 1
= 2
=2
=2
=2
= 2
= 2
= 2
= 2
6. If we trim off the Oth column through the nth column, inclusive,
the sequence of numbers obtained by altemately subtracting and
adding is the same as the nth column.
Co
1 I
1 / 1
1 I 2- 1
/
1
1 /
/
1/ 7
1/ 8 -
113-3+1
114- 6+4- 1
/ 5- 10 + 10- 5 + 1
6 - 15 + 20 - 15 + 6 - 1
- 21 + 35- 35 + 21 - 7 -+ 1
28 + 56 - 70 + 56- 28 + 8 - 1
104
!
'
=1
=1
=1
=1
=1
=1
=1
=1
\
1
1
1
c,
1/
2 I 1
313- 1
1
4j
6- 4+ 1
1
5 I 10- 10 + 5 - 1
6 15- 20 + 15 - 6 + 1
1
1
7 ~ 1 - 35 + 35- 21 + 7 - 1
8 I 28 - 56 + 70- 56+ 28- 8 + 1
1
1
= 1
=2
=3
=4
=5
=6
=7
1
1
1
C:a..
1 /
1
3
3 /1
6 I 4- 1
4
1
10 /10- 5 + 1
1
5
1
6 15 I 20 - 15 + 6 - 1
1
7 21 I 35 - 35 T 21 - 7 + 1
1
8
28/ 56 - 70 + 56- 28 + 8 - 1
1
2
105
= 1
=3
=6
= 10
= 15
= 21
@
7. For any row n, if we call the elements ao. a 1, a2, ... , an, then for any
row, except the Oth row, the sum of the even numbered elements (ao.
a2, a 4 , etc.) is the same as the sum of the odd numbered elements (a 1,
a 3 , as. etc.). The sum in each case is 2n-1.
1
1
1._
1
1
1
~
_!_
.§_ 4
3
4
1...
1
2
10 5
1
_!_ 6 15 20 lli 6
l.
1
7 21 35 35 21 .J_ 1
1
8 28 56 70 56 28 8
.1...
1
10
5
= 1
=2
=4
=8
= 16
= 32
= 64
= 128
=20
=21
=22
=23
=24
=25
=26
=27
= 1
=2
=4
=8
= 16
= 32
= 64
= 128
=20
=21
=22
=23
=24
=25
=26
=27
1
_1_
1
1
1
1
..1_
~
4
3
6
1
_L
..1..
1
-
10 5
1
1 .§_ 15 20 15 _§_ 1
1 ..]_ 21 35 35 21 7 ...L
1
1L 28 56 70 56 28 1L 1
1
5
10
106
'
il
8. Starting with any element in any row and working to the left,
altemately subtracting and adding, the result will be the number
directly above and to the right of the original number chosen.
1
1
1
1
1
11
1
1
2
3
4
6~1
4
1
(s7e
10
10
1
6 15 20 1.§_ 6
1
11
7
1
1
7 21 35 35 21
1
1
8 28 56 70 56 28 8
5
6-4+1=3
15 - 20 + 15 - 6 + 1 = 5
9. In any row, altemately subtracting and adding as far as some
element ak gives
(i) ak of the previous row if k is even
(ii) - ak of the previous row if k is odd.
1
1
1
2
1
327
1
1
1
4
11 - 4 + 6
1
5 10 10 5
1
1
1
6 1~/-%15 6
11 .. 7 + 21 - 35:' 35 21
7
1
1
1
8 28 56 70 56 28 8
107
'
10. If the elements in row n are squared and these squares are added,
the result is the middle element in row (2n). (This sequence of
numbers is the vertical line through the middle of the triangle.)
(iJ
=1
1,1. + 12.
.2.
la,·& +a. 12..
2..
2,;
l+3-t3+1
1"' + 4 a. i"(§f+ 4 a.. -t 1a.
4
1 + 5 4 1' 102,:+ 102:. Sot+ 1.t.
1 a. + 6 2.+ 15.:I.-t @'\ 15 "+ 6 a.+ 12..
4
1 + 7a+21a.+35\35'+21a._,.. 7~+ 12..
1 a. + 8 ~ 28 \ 56 4.+
56~ 28 ~ 8 .a.+ 12..
®\
=2
=6
= 20
= 70
= 252
= 924
= 3432
= 12,870
11. If a row has an even number of elements, then squaring the
elements and alternately subtracting and adding gives a result of zero.
1
1.2.- 1;t.
1
r~·-
2
3 ~+
=0
1
3'2.-
1~
=0
1
4
6
4
1
2.
.l.
.2....
;z,..
2...
1 - 5 + 10- 10 T 5 - 1
1
1
6 15 20 15 6
:I.
1
8
28
56
70
56
28
=0
8
1
12. If a row, n, has an odd number of elements, then squaring the
elements and alternately subtracting and adding gives
(i) the middle element of the row if
108
n
2 is even
(ii) the negative of the middle element of the row if
1
1
:a..
1
.a.
1- 4
3
a.
.t.
6- 4 + 14
1
1
5 10 10 5
.t.
.z.
~
2..
a.
1,1.,- 6 + 15 - 20 + 15 - 6 + 1~
1
1
7
8
= 1
0.
: 2 1s even
= -2
:~is
=6
:i
1
3
.t.
t
2 is odd.
1
.a..
1 - 2 + 1.;1.
.t.
n
= -20 :
~
odd
is even
is odd
21 35 35 21 7
1
28 56 70 56 28 8
1
13. We define a "string" as the sequence of numbers in any row and
"string multiplication" as the inner product of two strings.
For example:
r3 . r4 = (1, 3, 3, 1, 0) . (1, 4, 6, 4, 1)
= 1·1 + 3·4 + 3·6 + 1·4 + 0·1
= 35
(We annex zeros to the right of the shorter string to make the two
strings of equal length.) 35 is the element in the seventh row and
third column. We write 35 = (;). 35 is also the element in the
. seventh row and fourth column and we may write 35 =
(~).
In general, if we string multiply the nth and mth rows we get
the element in the (n + m)th row and the nth column (or the mth
column.
What this says is that there are two ways to express each term of
the Pascal triangle (except for the middle term of any row). If an
element is in the (n + m)th row and nth column, it is also in the
109
(n + m)th row and mth column. We can "see" this from the symmetry
of the triangle. We can prove it using the explicit formula from
chapter 6,
(~) = r!(:~ r)!.
Proof:
(n~m)
=
(n+m)!
n!(n+m-n)!
=
(n+m)!
n!(m)!
=
(n+m)!
m!(n+m-m)!
(n~)
=
Now, since rn · rm represents the element in the (n + m)th row and
nth column as well as the element in the (n + m)th row and mth
column, we have rn · rm = rm · rn, i.e. string multiplication is
commutative.
14. For any column, n, the sum of the elements a 0 through ak,
inclusive, is the same as the element ak in the (n + l)st column.
Cz.
1
1
1
c.r
1
1
1
8
8
1
1
1
1
The shape of these figures resemble hockey sticks and so we call
this the hockey stick property.
11 0
15. For any row, n, which contains an odd number of elements, the
as~}
middle element may be represented
i.e. the middle element Is
in row n and column ~.
=(g)~ ro.
G)
1
1
3
1
1
1
8
21
28
35
56
1
4
=(~) ~ r4. c2
1
10
@
15
6
7
®
10
5
=G)~ r2, c~
1
3
4
1
1
1
®
1
co
5
1
15
35 21
=(~)~ r6, c3
1
6
1
7
®56 28
8
1
=(~)~ r8, c4
16. For any parallelogram drawn which has one vertex at the apex of
the triangle, part of do acting as one side and part of c 0 acting as one
side: the sum of the numbers inside is equal to one less than the
element two rows down and pointed to by the lower corner of the
parallelogram.
1
2
1
3
4
6
4
1
1
5 10 10 5
1
1
6@ 20 15 6
1
7 21 35 35 21 7
1
1
8 28 56 70 56 28 8
1
1
111
1
1
1
1
1
1
1
1
8
8
112
1
p
Hexagons
1. If a hexagon is drawn around a group of seven elements as shown,
the sum of the numbers is twice the number directly below the
hexagon. The six numbers that surround the middle numb~r is called
a "ring."
1
1
1
1
1
3
1
1
1
1
1
6
8
7
1
1
28 8
1
1
1
1
1 13
•
2. For any hexagon drawn as above, we consider the six numbers that
make up the ring. The product of every other number is equal to the
product of the remaining three numbers.
1
1
1
1
1
1
1
1
1
6
7
1
1
8
1
5
15
6
21
28 56
1
1
8
1
3·4·10=3·10·4
3. The product of the six numbers in any ring is a perfect square.
If we label the six elements as such:
property 2 shows us that ade
= bfc.
114
Therefore, (ade)(bfc)
= (ade)2.
1
1
1
1
1
1
1
1
6
1
1
1
1
1
8
8
1
42. 32. 102 = 14,400 = 1202
6. 4. 5. 15. 20. 10
115
= 360,000 = 6002
Fibonacci Sequences
The sequence of numbers 1, 1, 2, 3, 5, 8, 13, 21 is part of the
sequence known as Fibonacci numbers. Although these numbers were
largely ignored for hundreds of years after the publication of Liber
Abaci by Leonardo Fibonacci in Pisa in 1202, there now exists
extensive literature on Fibonacci numbers and their many
ramifications in nature ~d mathematics. Pascal apparently did not
know that the Fibonacci series was embedded in the triangle; it seems
not to have been noticed until late in the 19th century.
Each number in the Fibonacci sequence is the sum of the two
numbers preceding it and the limit of the sequence formed by
dividing each element by its successor is known as the golden section
(approximately 0.618 ... ).
In the following diagram, the underlined groups of numbers are
added together to form a Fibonacci number. Each group of numbers in
the triangle formed this way is called a Fibonacci diagonal.
Diagonal Sums
1
1
2
3
1
1
70
56
7
28
i i 6
1
8
1
5
8
13
21
34
We now take the Fibonacci numbers and consider the sums of
these:
1,
1,
F3
F4
F5
F6
F7
F8
Fg
2,
3,
5,
8,
13,
21,
34, ...
F1 = 1
F1 + F2 = 2
F 1 + F 2 + F3 = 4
F 1 + F2 + F3 + F4 =7
F 1 + F 2 + F 3 + F 4 +F5 = 12
F 1 + F 2 + F 3 + F 4 +F5 + F 6 = 20
F 1 + F 2 + F 3 + F 4 +F5 + F 6 + F 7 = 33
and so on.
We can find these partial sums from Pascal's triangle by trimming
off the Oth column and taking the sums of the following diagonals:
Diagonal Sums
Co
1
2
4
7
12
20
33
1
6
1
8
28
35 35 21 7
56 70 56 28
117
1
1
8
1
If we take our first set of partial sums and add these numbers we
get new partial sums.
1,
4,
7,
Fs
F6
F7
F8
F9
12,
20,
33,
54,
88, ...
F1 = 1
F1 + F2 = 3
F1 + F2 + F3 = 7
F1 + F2 + F3 + F 4 = 14
F 1 + F 2 + F 3 + F 4 +Fs = 26
F 1 + F 2 + F 3 + F 4 +Fs + F 6 = 46
F 1 + F 2 + F 3 + F 4 +Fs + F 6 + F 7 = 79
and so on.
We can find these partial sums from Pascal's triangle by
trimming off the Oth column and first column and taking the sums of
the following diagonals:
Diagonal Sums
1
1
3
7
14
26
46
79
1
1
1
1
1
1
8
1
In general, if k columns are trimmed off the triangle, the
Fibonacci diagonals give the k-fold partial sums of the Fibonacci series.
118
Primes
1. If n is prime then n is a factor of every number in row n
except the ones.
1
1
1
2
1
1
1
1
3
3
6
4
2
3
1
1
4,
1
10 5
1
1
6 15 20 15 6
7
1
7
21 35 35 21
1
28 56 70 56 28 8
1
8
1
5
10
5
7
1
2. A sieve for the prime numbers can be obtained by a simple
manipulation of the Pascal triangle. We shift then+ 1 entries of row n
to occupy places in the columns 2n to 3n, inclusive. Also, the entries
in row n which are divisible by n are circled.
~
0
0
1
1
1
2
3
1
1
2
4
5
6
1
2
1
1
3
4
7
8
9 10 11 12 13
3
J
1
1
4
5
6
4
1
1
5
10 lD
6
1
6
14 15 16 17 18 19
5
8
1
7 21 35 35 21
1
8
28 56
1
9
10
-
1
15 20 15 6
1
7
'"
I.
-------
-
-
[Source: 22, p. 3]
119
-
1-
... ...
---
9
--- --
Now the natural number k is a prime number if and only if all the
numbers in column k are circled. In our chart we see that 2, 3, 5, 7,
11, 13, 17, and 19 are prime.
Proof:
The table verifies the claim fork= 1 through 19. For
k = 2in, an even number greater than two, we have m > 1,
and also that the first entry of row m occurs in column k.
But this entry is a 1, and it will not be circled because, for
m > 1, m does not divide 1. Since even numbers
exceeding 2 are composite, the hypothesis is true for k
even.
Suppose, then, that k is odd. We shall show that if k is a
prime number p, then every entry in column k is circled,
and that if k is composite, then at least one number in
column k is uncircled. We observe that the rows, n, which
put entries in column k are those for which
2n ::; k ::; 3n,
k
k
or
n ::; 2 and ::; n
3
or
k
k
-<n<3- - 2'
A look at row n in the next chart shows that the entry it
puts in column k is
row\col ---
2n
(k-~n}
2n + 1 2n + 2 ---
k
---
---
---
---
---
---
---
n
---
(~)
(~)
(~)
---
(k~n)
---
---
---
---
---
--- ---
120
--- 3n ----- --- ----- (~)
---
--- ---
(i) If k = p, a prime exceeding 3, the entries in column k
are
(p~nJ
where
Since p exceeds 3, we have 1 < n < p, implying that n and
p are relatively prime numbers. This means that n and
p- 2n are relatively prime.
Now, for each of the entries
(p~n) in such a column, we
have
n!
n )
( p- 2n = (p - 2n)!(n - (p - 2n))!
n(n- 1)!
= (p -2n)(p - 2n - 1)!(3n - p)!
n
(n- 1)!
= p - 2n · (p - 2n - l)![(n - 1) - (p - 2n - 1)]!
n
(
n-1
= p - 2n p - 2n - 1
)
giving:
We see that n divides the left side of this equation. But,
since n and p - 2n are relatively prime, we conclude that n
n}
and the entry is therefore circled.
divides (p ~2
(ii) Finally, suppose that k is an odd composite number.
As such, it is the product of two or more odd primes. Let
p denote an odd prime divisor of k and let k = p(2r + 1).
(p will divide into k an odd number of times.) Since k is
121
composite, we must have r :::: 1. Consequently, we have
p ~ pr, and
2pr < k = 2pr + p
~
3pr
Thus, now n = pr contributes to column k, and its entry
there is
We shall show that n = pr does not divide this number,
implying that column k has an uncircled number.
Attempting the division, we have
..!_ (pr) _ ..!_. pr(pr- 1)(pr- 2) · · · (pr- p + 1)
pr P - pr
1·2 ·3 · · ·p
= (pr - 1)(pr -2) · · · [pr - (p - 1)]
1·2·3···p
No factor (pr- i) in the numerator is divisible by the prime
p because 1 ~ i ~ p - 1. Since the prime p occurs in the
denominator, the fraction does not reduce to an integer,
showing that n does not divide the entry.
The proof is complete.
122
3. We now look at the prime number 11 because of the striking
information the triangle provides in giving powers of 11. The first five
powers of 11 are given by the first five rows:
= 110
= Ill
= 112
= 113
= 114
1
J,
1
1
1
1
1
2
3
3
4
6
1
4
1
1
10 5
1
1
6 15 20 15 6
1
7 21 35 35 21 7
1
1
8 28 56 70 56 28 8
1
5
10
1
Now, the fifth row sho~ld'b~ 115 = 161,051, but it is not. We observe,
however, that this is the first>r:ow with two-digit numbers. If we
interpret each number as indic~h~ga multiple of the place value of
that spot in decimal notation,. th~ flfth row can be interpreted
(reading right to left) as (lxl) + (5xl,b) + (IOxlOO) + (IOxlOOO) +
.'(5xl0,000) + (lxlOO,OOO); which gives the.correct value of 115.
Interpreted this way, each nth row is 11n.
123
Multiples
1. There is a rule for determining how many odd numbers are in any
given row of the triangle:
(I) Express n in binary notation
(2) Count the number of ones in the notation
(3) Raise 2 to the power equal to the number found in
step 2.
examples:
RowS:
1
5
10
10 5
1
5 = 1012
There are two 1s.
22 = 4 = the number of odd numbers in row 5.
Row7:
1
7
21
35
35
21
7
1
7 = 1112
There are three 1s.
23 = 8 = the number of odd numbers in row 7.
2. We have three possibilities for adding two numbers in terms of
even and odd.
Even + Even = Even
Odd + Odd = Even
Even + Odd = Odd
If we shade the even numbers in Pascal's triangle and translate
the above rules into a geometric pattern, our work is easily done. We
124
know that since all the first and last entries of each row are 1s that
these will be unshaded. We may work down the triangle quickly
without even knowing the numbers. Two unshaded numbers will
produce a shaded number beneath. Two shaded numbers will produce
a shaded number beneath. One shaded and one unshaded number will
produce an unshaded number beneath.
We look at the first 64 rows and observe the symmetrical
pattem created upon shading the even numbers.
Even Numbers: Multiples of 2
[Blank grid source: 34, p. 77]
125
We investigate the procedure to "count" the number of even
numbers and the number of odd numbers in the first n rows of the
triangle.
We can fmd the total number of entries in any triangular array by
taking one-half the product of the number of rows and the number
that is one more than the number of rows.
1st 4 rows:
4·5
2
= 10 numbers
1st 8 rows:
8·9
2
= 36 numbers
1st 16 rows:
16. 17
= 136 numbers
2
1st 32 rows:
32.33
= 528 numbers
2
1st 64 rows:
64.65
= 2080 numbers
2
The first four rows have 9 odd numbers (and 10 - 9 = 1 even)
The first eight rows contain three of the patterns found in the
first four rows. The middle section is all even. Therefore, the first
eight rows have 3 x 9 = 27 odd numbers (and 36 - 27 = 9 even
numbers).
The first 16 rows contain three of the patterns found in the first
eight rows. The middle section is all even. Therefore, the first
sixteen rows have 3 x 27 = 81 odd numbers (and 136 -81 =55 even
numbers).
The first thirty-two rows contain three of the patterns found in
the first sixteen rows. The middle section is all even. Therefore, the
first thirty-two rows have 3 x 81 = 243 odd numbers (and 528 - 243 =
285 even numbers).
126
Generalizing, we have:
#rows
#entries
# odd entries
# even entries
3n
3. We take a moment here to discuss "perfect" numbers. A perfect
number is a composite number that is equal to the sum of all its
divisors, excluding itself. The smallest perfect number is 6 where
6 = 1 + 2 + 3. The next perfect number is 28 = 1 + 2 + 4 + 7 + 14.
The next three are 496, 8128, and 33,550,336. It is not known if
there is a largest perfect number or if the list is infinite. [16, p. 130]
Euclid proved that every number of the form 2n-1 (2n - 1), where
2n - 1 is a prime (called Mersenne primes), is an even perfect
number. Euler showed that all even perfect numbers are of the form
1
p(p + ), where p is a Mersenne prime. This expression is also the
2
formula for a triangular number.
Our triangle has a series of shaded (even) triangles down the
center. The formula for the number of entries in the nth central
triangle, moving down from the apex, is 2n-1(2n- 1), the formula for
perfect numbers. All even perfect numbers appear in the pattern,
therefore, as the number of entries in the nth central triangle
wherever 2n-1 is prime. The first four central triangles contain 6, 28,
120, 496, and 2016 entries. Three of these numbers, 6, 28, and 496,
are perfect.
4. On the following pages is a series of triangles with the multiples of
three through ten, inclusive, shaded. They are included for the
beautiful visual symmetry provided. The number of rows considered
for each multiple varies as the patterns determine this. Note that the
percentage of the entries shaded increases as the number of rows
plotted increases.
127
Multi:gle
# rows in :gattem
54
64
75
64
49
64
54
64
3
4
5
6
7
8
9
10
128
Multiples of 3
129
Multiples of 4
130
Multiples of 5
131
'
'
Multiples of 6
132
il
Multiples of 7
133
.
Multiples of 8
134
Multiples of 9
135
Multiples of 10
136
Figurate Numbers
Numbers that can be represented in geometric form are called
figurate or polygonal numbers. The numbers 1, 3, 6, and 10 are
examples of triangular numbers, since each of these counts the
number of dots that can be arranged evenly in an equilateral triangle.
The following diagrams show that the first four square numbers are 1,
4, 9, 16; and the first four pentagonal numbers are 1, 5, 12, 22.
Triangular
1
Square
•
1
3
4
6
9
•
D- --
--
16
10
•
137
...
41.
••
4•
Pentagonal
•
1
5
12
22
We now look closely at how Pascal set up his famous triangle
with regard to figurate numbers. (We use the triangle in the form
found in Pascal's treatise.)
Pascal referred to each horizontal row as a "rang parallele." In
these rows we have figurate numbers. Pascal distinguishes them into
orders. In the first rang parallele there is a series of simple units 1,
1, 1, 1, ... , which Pascal calls numbers of the first order; he calls the
numbers 1, 2, 3, 4, ... , which form the second rang parallele, numbers
of the second order. These are the natural numbers, whose
differences are the numbers of the first series. In the third row we
have numbers of the third order, 1, 3, 6, 10, ... , which had already
received the -name of triangular numbers. The differences of these
numbers are the members of the second series. The numbers of the
fourth order, 1, 4, 10, 20, ... , are the pyramidal numbers whose
differences are the triangular numbers. Pascal says that the numbers
138
of the fifth order, 1, 5, 15, 35, ... , had not yet received an express
name, and he proposes to call them triangulo-trtangulaires whose
differences are the pyramidal numbers, and so on.
The vertical columns are called rang perpendiculaires and are
built on the same principle.
1st order
(simple units)
1
1
1
1
2nd order
(natural)
1
3rd order
(triangular)
1
3
6
10
15
21
28
4th order
(pyramidal)
1
4
10 20
35
56
84
5th order
(triangulotrtangulairal
or pentatopal)
1
5
15
70 126
1
1
1
1
1
1
!111!11111M11
2
4
3
5
6
7
8
9
1111!1/1MI1
11!1M/111
11111111/l
35
36
Each sequence of numbers is called a "difference sequence" for
the sequence of numbers below. The row of simple units gives us a
sequence of "common differences" (since all the numbers are 1). Any
sequence obtained from any row will lead to this common sequence in
a finite number of steps. This process is called "finite differentiation."
Likewise, the process of starting with a sequence and arriving at a new
sequence by adding successive terms is called "integration."
139
@
From the following chart we can come up with recursive
formulas for the numbers of each of the three figurate classes listed
below. Notice the use ofTn, the nth triangular number.
n
Triangular T n
1
2
3
5
6
1
3
6
10
15
21
.
.
.
.
.
4
n
Tn=T(n-l)+n
Square Sn
Pentagonal Pn
1
1
5
12
22
35
51
4
9
16
25
36
.
.
.
Sn=2T(n-l)+n
Pn=3T(n-l)+n
General
Rule:
Tn=
n(n+l)
2
Sn= n2
Pn=
n(3n-1)
2
Some interesting facts:
1. Every natural number is either a triangular number or a sum
of two or three triangular numbers.
2. Every square number is the sum of two successive triangular
numbers.
3. Every natural number is either a square number or a sum of
two, three, or four square numbers.
4. Every pentagonal number is the sum of a triangular number
and a square number or is the sum of three triangular numbers.
The following diagrams show where these figurate numbers can
be found in Pascal's triangle.
140
•
Triangular
1. In the second column or diagonal.
1
1
1
8
1
2. The result of adding the following pairs of numbers:
4
1
1
1
1
7
8
10 ~..~..1~---"""--1
5
6
15
20 ~1r-~.....
35 35
21
28 56
70
56
1
3. The result of taking one-half of the following products:
1
1
1
1
1
141
4. The result of subtracting the smaller number from the larger
number in each of the following pairs:
1
1
1
1
1
1
7
1
1
6
1
1
7
8
28
8
1
5. The result of taking the determinant of the following groups of
numbers:
1
1
1
1
1
7
1
1
8
28
56
70
1
Square
1. The result of adding the following pairs of numbers:
1
1
1
1
1
1
1
8
8
142
1
2. The result of subtracting the smaller number from the larger
number in each of the following pairs:
1
\o~"\:
~
1
1
<\
J,6
4
1
5
1.0' 10 5
1
1
1
6 15 \ 2o~5 6
1
1
7
7 21 3~35 ,21
8
1
70
56 28
1
28 56
8
Pentagonal
The result of adding each set of the following groups of four numbers:
1
1
1
3
3
·--1~
-
1
~
4
1
10~ 10
5
1
5,-: 20 15 6
1
21~ 5 35 21 7 1
5•
1
6
7
1
1
14
1
1
1
8
70
56
56
28
8
1
1
1
1
1
1
1
1
3
4
r;-
--,
~ 10
4
1
1
5
,5
6r 15 ~ 20 15 6
1
1
21
7
7 (1J' '35 35
1
1
1
8
281./56 70 56 28
8
1
1
143
Higher Dimensional Figurate Numbers
A tetrahedron is a pyramid with a triangular base. It is a threedimensional figure. A tetrahedral number is the number of dots in the
form of a tetrahedron composed from layers of triangular numbers.
The first four tetrahedral numbers are shown below:
1
0
1+3=4
1 + 3 + 6 = 10
1 + 3 + 6 + 10 = 20
144
A square pyramid has a square base. A square pyramidal number
is the number of dots in the form of a square pyramid composed of
layers of square numbers. The first four square pyramidal numbers are
pictured below:
•
1
1 +4
=5
1 + 4 + 9 = 14
145
1 + 4 + 9 + 16 = 30
146
A pentagon pyramidal number is the number of dots in the form
of a pentagon pyramid composed of layers of pentagonal numbers. The
first four pentagon pyramidal numbers are pictured below:
•
1
1+5=6
1 + 5 + 12 = 18
147
1 + 5 + 12 + 22 = 40
148
From the following chart we can come up with recursive
formulas for the numbers of each of the three higher dimensional
figurate classes listed below. Once again, notice the use of Tn, the nth
triangular number.
Square Pyramid
Sn
Pentagonal Pyramid
Pn
n
Triangular
Tn
Tetrahedral
tn
1
2
3
4
5
6
1
3
6
10
15
21
1
4
10
20
35
56
1
5
14
30
55
91
.
.
.
.
n
Tn
tn=Tn+tn-1
Sn=Tn+2tn-1
Pn=Tn+3tn-1
Rule:
Tn=
n(n+1)
2
tn=
n(n+ 1)(n+2)
6
Sn=
n(n+1)(2n+1)
6
Pn=
n(n+ 1)(3n)
6
1
6
18
40
75
126
The following diagrams show where these figurate numbers can
be found in Pascal's triangle:
149
Tetrahedral
In the third column or diagonal:
1
1
1
1
1
6
7
1
1
7
8
1
8
1
Sguare Pyramidal
1. The result of performing the indicated operations for each set of
the following groups of three numbers:
1
1
1
2~
1
1
1~
4
6 - 4
1~5
6 15- 20 15
1
1
1
~
1
1
6
1
1
35 21
7
1
70 56 28 8
~
2. The result of adding the following pairs of numbers:
1
1
1
1
1
2
(1
3
6
4
1
1
5 1o,(1o/5
1
1
6 15 20 15 6
1
1
2"C"3s/Js 21
1
1
1
8 28 56 70 56 28 8
1
1
1
3
1
4
150
Pentagon Pyramidal
The result of multiplying the following pairs of numbers:
1
1
6
7
1
1
1
70
151
56
28
1
8
1
Chapter 10
APPLICATIONS
Just as there seem to be a countless number of patterns found in
the arithmetical triangle. there also seem to be a countless number of
applications of the triangle. This final chapter examines seven
applications. a sampling which is intended to show how diverse the
possibilities can be.
152
Simplexes
The elements of geometry are points, lines, planes, and space:
/7
/
•
Portion
of a Line
Point
Portion
of a Plane
Portion
of a Space
A point is zero-dimensional, a line is one-dimensional, a plane is
two-dimensional, and space is three-dimensional. Each of these
elements can be thought of as a space of a given dimension. Two
distinct points determine a line. Three distinct, noncollinear points
determine a plane. Four distinct, noncoplanar points determine
space.
The simplest figure that can be drawn in any one of these spaces
is called a "simplex." In zero-dimensional space, the point is a
simplex of order zero. In one-dimensional space, the line segment is
a simplex of order one and the point is a simplex of order zero. The
triangle is a simplex of order two in two-dimensional space. The line
segment and the point exist in two-dimensional space as simplexes of
orders one and zero, respectively. The tetrahedron is a simplex of
order three in three-dimensional space. The point, line segment, and
triangle all exist as simplexes of orders zero, one, and two,
respectively, in three-dimensional space .
•
Point
•
•
Line segment
Triangle
153
Tetrahedron
Each simplex of a given order is composed of elements that are
simplexes of lesser orders. The composition of each simplex of a
given order is given in the following table.
Simplex
Point
Line segment
Triangle
Tetrahedron
Subspaces
Points Line segments Triangles
(order 0)
(order 1)
(order 2)
1
2
3
0
0
1
0
1
3
6
4
Tetrahedrons
(order 3)
0
0
0
1
4
Except for the leading 1, each row of Pascal's triangle gives the
number of subspaces contained in each simplex of order equal to the
row number minus one.
1 /2
r 1:
r2:
r3:
r4 :
1
1 /3
3
1
1 /4
4
1
6
1/ 5 10 10 5
1
1
1/ 6 15 20 15 6
1
1/ 7 21 35 35 21 7
1 / 8 28 56 70 56 28 8
order
order
order
order
1
2
3
4
-
1
1
1
1
=0
=1
=2
=3
1
On the assumption that four-dimensional space can be defined,
we can predict the number of subelements a simplex of order four
would have. The simplex of order four is called a "pentatope." A
pentatopal figure is pictured below:
154
We extend our chart to include order four and again may use
Pascal's triangle.
Simplex
Subspaces
Line
segments
(order 0) (order 1)
Points
Point
Line segment
Triangle
Tetrahedron
Pentatope
1
2
3
4
5
Triangles
TetraPentatopes
hedrons
(order 3)
(order 4)
(order 2)
0
1
3
6
10
0
0
1
4
10
1 y1
1
1/2
1
3
3
1
1 /4
4
1
6
1
1 / 5 10 10 5
1
/. 6 15 20 15 6
7 21 35 35 21 7
1
1
1
8 28 56 70 56 28 8
I
155
0
0
0
1
5
r1:
r2:
r3:
r4:
r5:
1
order
order
order
order
order
0
0
0
0
1
12 345-
1
1
1
1
1
=0
=1
=2
=3
=4
Dividing Space
We can divide a line with a point, a plane with a line, and space
with a plane.
Given a line, we seek the number of divisions made by n points.
Given a plane, we seek the number of divisions made by n lines. Given
space, we seek the number of divisions made by n planes. In each
case we look for the "largest" number of possible divisions. Therefore,
we want the lines non-parallel and no more than two lines passing
through the same point. We want the planes non-parallel, no more
than two planes containing the same line, and no more than three
planes containing the same point.
Points dividing a line:
Number of points
Number of divisions
0
(
1
<
2
(
3
(
4
<•
156
•
• •
• • •
• • •
>
1
)
2
>
3
)
4
~
5
Lines dividing a plane:
Number of divisions
Number of lines
0
L7
1
1
2
2
4
3
7
157
4
11
Planes dividing space:
Number of planes
Number of divisions
0
1
1
2
2
4
158
3
8
4 This is difficult to draw so we look for an analogy
with two dimensional space.
We repeat our case of dividing a plane with three lines and count the
number of regions in a new way to get seven regions as before.
interior triangle = 1 piece
vertices = 3 pieces
edges = 3 pieces
total = 7 pieces
Now we consider dividing space with four planes and count the
regions in a similar way.
interior tetrahedron = 1 piece
vertices = 4 pieces
edges = 6 pieces
faces = 4 pieces
total = 15 pieces
159
Now, it is easy to see that n points divide a line into n + l or
(g)+(~) parts. We now try to fmd a formula for the number of parts n
lines divide a plane into. We will then find a formula for the number of
parts n planes divide space into. Since no two of the lines dividing a
plane are taken as parallel, it is possible to find a circle sufficiently
large to contain all the points determined by intersections of pairs of
lines. Let us call each point that is the intersection of a pair of lines a
"derived point" and each point common to the circle and a line a
"boundary point." Then these boundary points determine a convex
polygon as shown. (We use four lines.)
Ia
l4
/3
0: derived points
(lines intersect here)
x: boundary points
(lines and circle
intersect here)
This figure consists of vertices (derived points and boundary
points), edges,. and regions.
l. To find V, the number of vertices: Each pair of lines
determines a unique derived point; there are (~)such points (two
lines taken at a time out of n lines). Each line determines two
boundary points; there are 2n such points (each line intersects the
circle in two places). Hence V
= (~) + 2n.
2. To find E, the number of edges: Consider the number of
edges on each vertex. On each derived point there are four edges, and
on each boundary point there are three edges. But counting the
160
number of edges by considering the vertices counts each edge twice
(since each edge contains two vertices.) Hence
2E =
4(~)
+ 3(2n)
3. To find R, the number of regions: By Euler's formula,
V - E + R = 1. We obtain R = 1 - V + E
= 1-
(~)-
= 1+n +
2n +
2(~) + 3n
(~)
=(g)+(~)+(~)
For our example, n
= 4, we
get
(6) + (i) + (~)
=1+4+6
R =
= 11
which is in agreement with the number of regions in our figure.
Since the number of regions within the polygon is the same as
the number of regions into which the lines partition the plane, we
have found the maximum number of divisions determined by n lines in
a plane:
(g)+(~)+(~).
We next consider the problem of finding the maximum number
of parts into which n planes partition space. It would seem that there
are
(g)+(~)+(~)+(~) parts.
We set out to prove this.
Proof: From our previous discussion we saw that one plane gives two
parts or
(b) +(~) +(~) +(~) =1+1+0 +0; two planes give
161
four parts or
fo) +~)+G)+~)=
give eight parts or (g)+ (I)+
four planes give 15 parts or
4. (Recall
(~) + (~) = 1 + 3
+ 3 + 1; and
(6) + (i) + (~) + (~) = 1 + 4 + 6 +
that(~)= 0 when
n planes give
1 + 2 + 1 + 0; three planes
r>
n.) We would like to show that
(g)+(~)+(~)+(~) parts by the use of induction.
Assume that the formula works for n = N, that is N planes
partition space into
(~) + (~) + (~) + (~)parts.
Introduce the
(N + 1)st plane. Since no planes are parallel, the (N + 1)st
plane intersects the previous N planes in N lines, which being
neither parallel nor concurrent, determine the maximum
number of parts of the (N + 1)st plane that can be determined
by N lines. The maximum number of parts
is(~)+(~)+(~) a
fact from our previous discussion. Each of these portions on
the (N + 1)st plane divides a previous region in threedimensional space into two regions; thus two new regions are
added for each portion while one region is lost.
We stop here to illustrate with the example of three lines
dividing a plane into seven parts:
162
The fourth line will cut each of the three lines into 4 - 2 = 2
segments and 2 rays. producing four additional regions for a
total of 7 + 4 = 11 regions.
Each new segment and each new ray will divide a previous
region into two regions. Thus, the (N + 1)st line creates N + 1
additional regions. (Our fourth line created four new regions.)
Thus, the total number of regions determined by N + 1 planes
is
(~) + (~) + (~) + (~) + (~) + (~) + (~)
=
(N; 1)
+
(N; 1)
+
(~)
+
(N; 1).
using Pascal's rule,
=
(N; 1) + (N; 1) + (N~ 1) + (N; 1}
since
=
(N~ 1) + (N; 1) + (N; 1) + (N; 1}
which is the desired result.
163
(g) = 1 for any n,
We now put our results in chart form and look for a pattern in
Pascal's triangle.
Number of pieces formed when
space is divided.
Number of
dividing elements
0
1
2
3
Line
by points
1
Plane
by lines
1
2
3
2
4
4
4
7
4
5
6
11
16
8
15
26
.
.
.
.
.
.
.
.
n
n+1
1+n+(~)
5
.
=(g}(~)
Space
by planes
1
2
=(g}(~}(~) (g}(~ }(~}(~)
We can find the numbers obtained from dividing a line by points in the
first column or diagonal of the triangle:
d,
1
1
70
164
56
1
or by adding the following pairs of numbers:
1
1
1
6
1
7
1
1
70 56 28 8
We can find the numbers obtained from dividing a plane by lines by
adding the following sets of numbers:
1
6
1
7
70 56 28
1
8
1
We can find the numbers obtained from dividing space by planes by
adding the following sets of numbers:
1
6
1
7
7
1
8
8
165
1
I
Polygons Inscribed In a Circle
Given a circle with n evenly spaced points on the circle, we can
inscribe a regular polygon with n sides. For n = 1, 2, 3, ... 10 the
following chart counts the number of points on the circle, the number
of line segments formed whose endpoints are on the circle, and the
number of each type of polygon formed whose vertices lie on the
circle. For each n the chart also gives the nth row of Pascal's triangle
with the deletion of the left-hand one.
166
'
CD~
~
@
•
0
I
1.(641
S 10 10 S I
b IS :;p
7 .U 3S
6
I
35~1
7
!5
•
~ .18 s' 70.56
•
/0 'IS 1.20 l!O
I
:1.!' 8
I
.1!.1.~/o 1.10 'IS t•
167
I
(Source: 18, p. 223]
Diagonals of Polygons
We next consider the number of diagonals in polygons.
Number of sides
of a polygon
Number
of diagonals·
Number of sides plus
number of diagonals
3
0
3
4
2
6
5
5
10
6
9
15
168
7
14
21
8
20
28
The number of diagonals is found in Pascal's triangle by adding
the following pairs of numbers:
1
1
6
1
1
7
1
8
1
169
1
The sums of the sides and diagonals are triangular numbers and
are found in the second diagonal or column:
1
1
1
1
1
8
1
170
1
Trigonometry
Pascal's triangle also occurs in trigonometry. We will examine
two sequences of trigonometric identities.
1. Tan (n8)
We consider the formula:
tan (A+ B)
For n = 1,
=
tan 8 =
For n
= 2,
tan28
For n
= 3,
tan 38
tan A+ tan B
1- tan A· tanB
1 tan 8
1
= tan
= tan
(8 + 8)
2 tan 8
=
. 1-tan2 8
(28 + 8)
tan 28 +tan 8
=----1 - tan 28 . tan 8
2 tan 8
-1---t-an---=-2-9 + tan 8
=-------2 tan8
1 - 1- tan28. tan 8
2 tan 8 + tan 8 - tan38
1- tan28
=---....,...---..,...__1 - tan28 - 2 tan28
1- tan28
3 tan 8 - tan38
1- 3 tan28
=----~-
Continuing, we look at the results in chart form. The terms are
written so that the powers of the terms of tan(n8) increase. On the
171
right we list the coefficients of the terms with the coefficients of the
numerators raised slightly. If coefficients are selected alternately from
denominator to numerator, the coefficients give us Pascal's triangle.
tan(ne)
n
Coefficients
1
1tane
1
1 1
2
2tane
1-tan2e
1
2 1
3
3tane-tan3e
1-3tan2e
1
3 3
1
4
4tan8-4tan3e
1-6tan2e+tan4e
1
4 6
4 1
5
5tan8-10tan3e+tan5e
1-10tan28+5tan4e
1
5 10 10 5
6
6tan8-20tan38+6tan 5e
1-15tan2e+ 15tan4e-tan6e
1
6 15 20 15
1
6 1
Note that the numerators contain odd powers and the
denominators contain even powers. Signs alternate in both the
numerator and denominator of each ratio. The largest power of any
term of each ratio is n. Our general equation is:
tan (ne)
=
(~) tane- (~)an3e + (~)an5e- ...
(g)- (~)an2e + (~)an4e- ...
172
2. cos(n9) + sin(n9)
We now consider the formulas
cos(A +B)= cosA cosB- sinA sinB and
sin(A + B) = sinA cosB + cosA sinB
For n = 1, cos(l9) = cose
sin(le) = sine
For n = 2, cos(29) = cos(e + 9)= cos2e - sin2e
sin(29) = sin(e + 9) = sinS cose + cose sinS
= 2sin9 cose
For n = 3, cos(39) = cos(29 + 9)
= cos29 cose - sin29 sine
= (cos2e - sm2e)cose - (2sin9 cose)sine
= cos3e - sin2e cose - 2sin2e cose
= cos3e - 3sin2e cose
sin(39) = sin(29 + e)
= sin29 cose + cos2e sine
= (2sin9 cose)cose + (cos2e - sin2e)sin9
= 2sin9 cos2e + sine cos2e - sin 3e
= 3sin9 cos2e - sin3e
173
Continuing, we look at the results in chart form;
n
cos(nS)
sin(nS)
1
cosS
sinS
2
cos2s - sin2s
2sinS cosS
3
cos3s - 3sm2s cosS
3sinS cos2s - sin3s
4
cos4s - 6sin2s cos2s
+ sin4s
4sinS cos3s - 4sin3s cosS
5
cos5s - 1Osin2s cos3s
+ 5sin4s cosS
5sinS cos4s - 1Osin3s cos2s
+ sin5s
Adding the two columns for cos(nS) and sin(nS) we obtain the
following chart:
n
cos(nS) + sin(nS)
1
cosS +sinS
2
cos2s + 2sinS cosS - sin2s
3
cos3s + 3sinS cos2s - 3sin2s cosS - sin3s
4
cos4s + 4sinS cos3s - 6sin2s cos2s - 4sin3s cosS + sin4s
5
cos5S+ 5sinS cos4s - 10sin2s cos3s - 10sin3s cos2s
+ 5sin4s cosS + sm5s
174
Paths
This basic problem begins with a rectangular grid. (Ours is
turned so we can use Pascal's triangle.) We label the points of
intersection. A path is called "simple" if it consists of one and only
one of the two following combinations of moves:
1. south-east and south-west
2. north-east and north-west.
We look for the number of different simple paths from point A to other
points.
A
The numbers in the grid are obtained easily by counting the
different simple paths from A to each of the indicated points. For
example, we see three simple paths from A to H: ABDH, ABEH, and
ACEH.
Other points may be reasoned logically as follows: consider the
number of paths from A to M. A path must pass through either H or I
176
just before it hits M. Since there are three different ways to get to H
and three different ways to get to I, there are 3 + 3 = 6 different
simple paths from AtoM. But, this is the rule that generates Pascal's
triangle. Hence, we may count all the simple paths starting from point
A by completing the grid in the same way we fill in the numbers in
Pascal's triangle.
177
The Apex Card Trick
And now it seems appropriate to come full circle and end with a
topic that had great influence on Pascal's treatise on the arithmetical
triangle--cards. The following is called the apex card trick.
A spectator is given a couple of card decks from which the face
cards and tens have been removed. He is asked to place any five cards
face up in a row. The dealer immediately finds a card in the deck that
he puts face down at a spot above the row as shown below.
D
In our example the spectator picks the five cards shown. The
ace is considered to have the value of one and the suits are not
involved. The spectator is now asked to build a pyramid of cards as
follows:
Each pair of cards in the row is added by the process of "casting
out nines." If the sum is above nine, nine is subtracted. This can be
done rapidly by adding the two digits in the sum. For example, the
first two cards in the bottom row of our example add up to 9 + 7 = 16.
Instead of subtracting nine from sixteen, the same result is obtained
by adding one and six. The sum is seven; therefore the spectator puts
a seven card above the first pair of cards. The second and third cards
add to 7 + 1 = 8, so an eight card is placed above them. This is
continued until a new row of four cards is obtained, and the procedure
is repeated until the pyramid reaches the face-down apex card. Our
example would finish up as shown:
178
D
When the apex card is turned over, it proves to be the correct
value for the final sum, 2 + 6 = 8. The trick can be done with any
number of cards in the initial row. The trick may of course be done
with pencil and paper instead of cards.
How is the apex card determined so quickly by the dealer? We
of course use Pascal's triangle to answer this question. Let's look at
the sums in our example without casting out nines. The pyramid
would be:
71
38 33
24 14 19
6
13
16 8
9
7
1
5
8
179
The bottom row corresponds to the row 1, 4, 6, 4, 1 in Pascal's
triangle since they both contain five entries. The ones, at the ends of
the fourth row of Pascal's triangle, tell us that, as we move upward
making additions, the value of each end card on the bottom row enters
only once into the final summation. That is because there is only one
path from each of these cards to the top. The fours that are second
from the ends tell us that the value of each card second from an end
enters four times into the final sum because there are four forking
paths from each of these cards to the top. The center six tells us that
there are six forking paths from the center card to the top, therefore
the value of this card enters six times into the fmal sum. Accordingly,
(1x9) + (4x7) + (6x1) + (4x5) + (1x8) = 71, the apex number. Since
this procedure gives the sum at the apex, it must also provide the apex
digital root if nines are cast out.
Magicians know the card trick as "Apex." It was orginated by a
German magician, Franz Braun, who published it about 1960 in his
regular column on mathematical tricks in "Magie," a German magic
periodical. [15, p. 194]
180
CONCLUSION
Pascal's triangle is truly an intriguing array of numbers. Not only
are there seemingly boundless patterns and relationships within the
triangular array itself, but also there are many mathematical topics in
which these numbers play a role.
Pascal states .I leave out many more [uses of the Arithmetic
Triangle] than I include; it is extraordinary how fertile in properties
this triangle is. Everyone can try his hand.
Towards the end of the seventeenth century it became the focal
point for the development of three branches of mathematics: the
study of infinite series, the calculus of finite differences and the theory
of probability.
One has to wonder, if indeed, Blaise Pascal himself was aware of
the powerful tool he created when he first set out to solve a gambling
problem!
II ••
II
181
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