Precalculus Honors
Mr. DeSalvo
2.1 Permutations
September 25, 2005
Suppose you are a contestant on a game show and are ushered into a room
where there are three closed doors. Your host indicates that behind one of the doors
is a large stack of yen. Behind each of the other two doors is a goat, for which you
really do not have much use. You have to select one door, and will receive whatever
“prize” is behind that door.
You then proceed to select a door, say door number one. However, before you
open that door, the host opens one of the other two doors and reveals, in all his
glory, a goat. You are then asked whether you would like to stay with your original
choice, or switch to the other unopened door.
The question is: Should you remain with your original choice? Should you
switch to the remaining door? Does it make a difference? What will you do with the
goat if you win it? The answer to this problem (not what to do with the goat, that
answer is obvious) has been vigorously debated by numerous mathematicians, yet it
does not even take any complicated mathematical knowledge to solve. What do you
think is the correct answer? In another handout we will answer this question, and
the answer may surprise you.
Now, in order to get to this answer we need to recall, or in some cases, learn for
the first time, how to count mathematically. This is much different than how you
learned to count on Sesame Street, so make sure you pay attention.
FUNDAMENTAL PRINCIPLE OF COUNTING
Suppose that a first task can be completed in m1 ways, a second task in m2 ways, and
so on until we reach the rth task that can be done in mr ways; then the total number of
ways in which these tasks can be completed together is the product
m1m2 L mr
EXAMPLE 1: A club consists of 15 boys and 20 girls. They wish to elect officers
! and a boy as vice-president. They also wish to elect
consisting of a girl as president
a treasurer and a secretary, who may be of either sex. How many sets of officers are
possible?
Solution: There are 20 choices for president and 15 choices for vice president.
Thereafter, since 2 club members have been chosen and the remaining positions can
be filled by members of either sex, 33 members are left for treasurer and 32 for
secretary, or vice versa. Then by the Fundamental Principle of Counting, the total
number of choices is
20 * 15 * 33 * 32 = 316, 800
EXERCISE 1: How many 3-digit whole numbers can be formed if zero is not an
acceptable digit in the hundreds place and repetition of digits are allowed? What if
repetition is not allowed?
DEFINITION OF PERMUTATION
A permutation of n distinct elements taken r at a time is an ordered arrangement,
without repetitions, of r of the n elements. The number of permutations of n elements
taken r at a time is denoted by nPr.
EXAMPLE 2: How many 3-letter “words” composed from the 26 letters of the
alphabet are possible? No duplication of letters is permitted.
Solution: Since duplication of letters is not permitted, once a letter is chosen, it may
not be selected again. Therefore, the first letter may be any of the 26 letters of the
alphabet, the second any of the remaining 25, and the third from the remaining 24.
Thus the total number of different “words” is 26 * 25 * 24 = 15,600. Since we have
taken 3 elements out of 26, without repetitions and in all possible orders, we say
there are 15,600 permutations.
EXERCISE 2: In how many ways can four coins (one quarter, one dime, one
nickel, and one penny) be arranged in a row?
PERMUTATION FORMULAS
For n distinct elements taken r at a time, where 1 ≤ r ≤ n:
n P r.
= n(n – 1)(n – 2) . . . ( n – r + 1)
n
Pr =
n!
(n " r)!
Using our first method 7 P4 =!7 * 6 * 5 * 4 = 840
7! 7 * 6 * 5 * 4 * 3!
=
Using the second method 7 P4 =
, the 3!’s cancel, and the
3!
(7 " 4)!
!
7! 7 * 6 * 5 * 4 * 3!
=
methods are equal, since 7 P4 =
= 7 * 6 * 5 * 4 = 840.
3!
(7 " 4)!
!
!
!
!
Please keep in mind when applying these formulas that the following must be true:
• The n elements must all be different.
• No element is repeated in a permutation.
• Order of elements is important.
EXAMPLE 3: A club consisting of 10 members wants to pick three officers, president,
a vice-president, and a secretary-treasurer. How many sets of officers are possible?
Solution:
10
P3 =
10!
10! 10 # 9 # 8 # 7/ !
=
=
= 720 sets.
(10 " 3)! 7!
7/ !
EXERCISE 3: A family consisting of 2 parents and 3 children are going to be arranged in
a row by a photographer. If the parents must be next to each other, how many
!
arrangements
are possible? (Hint: This example makes use of both permutations and the
fundamental counting principle.)
Permutations can also be formed using collections of objects not all of which
are distinct from one another. For example, in the word ELEMENT the 3 E’s are
not distinguishable. How many ways can these 7 letters be arranged to form all the
distinguishable permutations?
One such permutation is TENEMEL. If the E’s were momentarily made
distinguishable, by using subscripts, and the consonants remain fixed, then
TENEMEL produces these 3! = 6 permutations:
TE1NE2ME3L
TE2NE1ME3L
TE3NE1ME2L
TE1NE3ME2L
TE2NE3ME1L
TE3NE2ME1L
Similarly, every distinguishable permutation of the letters in ELEMENT produces
3! Permutations of the letters in E1LE2ME3NT, of which there are 7! = 5040. Then
letting P be the number of distinguishable permutations of the letters in ELEMENT,
5040 7!
we have 6P = 5040, or P =
= .
6
3!
This type of reasoning can be extended to produce the following result:
!
The number of distinguishable permutations of n objects of which n1 are of one kind,
n2 are of another kind, . . . , nk are of another kind is given by
n!
n1!n 2!Ln k!
!
EXAMPLE 4: Find the number of distinguishable permutations using all the letters
in each word.
(a) BOOK
(b) REFERRED
Solution:
(a) Use n1 = 2 , since there are two O’s, and n = 4 to get
(b) Use n1 =3, n2 = 3, and n = 8 to get
!
4!
= 12 .
2!
8!
= 1120 .
3! 3!
! permutations using all of the
EXERCISE 4: Find the number of distinguishable
letters in the word BEGINNING.
!
Homework Problems
1. How many ways can the manager of a baseball team select a pitcher and a
catcher for the game if he has 5 pitchers and 3 catchers?
For questions 2 – 5, consider three-letter “words” to be formed by using the vowels
a, e, i, o, and u.
2. How many different words can be formed if repetitions are not allowed? are
allowed?
3. How many different words without repetitions can be formed whose middle letter is o?
4. How many different words without repetitions can be formed whose letters at the
ends are u and i?
5. How many different words can be formed containing the letter a and 2 other
letters?
For problems 6 and 7 only consider three-digit numbers that can be formed using
the digits 1, 2, 3, 4, 5, 6, 7, 8, 9, and repetition is not allowed.
6. How many 3-digit numbers can be formed that are even?
7. How many 3-digit numbers can be formed that are greater than 600?
8. (a) In how many different ways can the letters of STUDY be arranged using
each letter only once in each arrangement?
(b) How many arrangements are there with S and T in the first 2 positions?
(c) How many arrangements with S and T next to each other? Not next to each other?
9. A license plate is formed by listing 2 letters of the alphabet followed by 3 digits.
How many different license plates are possible:
(a) If repetition of letters and digits are not allowed?
(b) If repetitions are allowed?
10. Solve for n: (a) n P1 = 10
(b) n P4 = 840
(c)
12
Pn = 8(12 Pn"1 )
11. A student has room for six books on a shelf. The books consist of six different textbooks
for six different subjects, two of which are Precalculus Honors and History.
!
!
!
(a) In how many ways can the books be arranged?
(b) In how many ways can they be arranged if the History book is in the 1st
position?
(c) In how many ways can they be arranged with the History and Precalculus
Honors books next to each other?
12. How many distinguishable permutations can be formed using all the letters in the
words
DELEGATE, COLLEGE, and STATEMENTS?