December 9, 2014
Any calculator permitted on N.Y.S. Regents examinations may be used.
The word “compute” calls for an exact answer in simplest form.
6 - Permutations and Combinations (No Probability)
1. Compute the number of distinct arrangements of the letters in the word
PERMUTATION.
2. There are nine men and eight women on the reality cooking show Cooking With Gas.
The host must choose two men and two women to participate in a challenge. Compute the
number of ways in which the host may do this.
3. Susie is writing the 5040 arrangements of the letters in ORANGES in alphabetical order:
AEGNORS, AEGNOSR, AEGNROS, . . . . What is the 2014th arrangement she will write?
December 2014
6 - Permutations and Combinations (No Probability)
1. Compute the number of distinct arrangements of the letters in the word
PERMUTATION.
SOLUTION: 19958400 The number of ways to arrange 11 letters is 11!. However, two
of the letters are T’s, so divide by 2! = 2 to obtain 11!
= 19958400.
2
2. There are nine men and eight women on the reality cooking show Cooking With Gas.
The host must choose two men and two women to participate in a challenge. Compute the
number of ways in which the host may do this.
9
= 36 ways
SOLUTION: 1008 The host chooses two of the nine men, and there are
2
8
to do this. The host chooses two of the eight women, and there are
= 28 ways to do
2
that. The answer is 36 · 28 = 1008.
3. Susie is writing the 5040 arrangements of the letters in ORANGES in alphabetical order:
AEGNORS, AEGNOSR, AEGNROS, . . . . What is the 2014th arrangement she will write?
SOLUTION: GROSENA Of the 5040 arrangements, the first 720 will start with A and
the next 720 will start with E, so the 2014th will start with G. Then, of those 720 (and we
need the 574th of those 720), the first 480 will begin with A or E or N or O, so the next
letter is R. Of the 120 arrangements that start with GR (and we need the 94th of those),
the first 72 will begin with A or E or N, so the next letter is O. Of the 24 arrangements
that begin with GRO (and we need the 22nd of those), the first 18 will begin with A or E
or N, so the next letter is S. The 6 arrangements that begin with GROS end in AEN, ANE,
EAN, ENA, NAE, and NEA. We need the fourth of those, so our answer is GROSENA.
Author: George Reuter ([email protected])
www.mathmeets.com
Reviewer: John Unson
December 3, 2013
Calculators are not permitted on this contest.
The word “compute” calls for an exact answer in simplest form.
6 - Permutations and Combinations (No Probability)
1. Ten people are in the audience of a game show. Four of them are invited to stand in a
line on stage. Compute the number of different ways this could occur.
2. Compute the number of permutations of BALLOON in which the two O’s are
side-by-side. For example, OOBALLN is one such arrangement, but BALLONO isn’t.
3. Let a, b, c, and d be non-negative integers. How many ordered 4-tuples exist such that
a + b + c + d = 11?
December 2013
6 - Permutations and Combinations (No Probability)
1. Ten people are in the audience of a game show. Four of them are invited to stand in a
line on stage. Compute the number of different ways this could occur.
SOLUTION: 5040 We have ten options for the first person on line, nine for the second,
and so on. Our answer is 10 · 9 · 8 · 7 = 5040.
2. Compute the number of permutations of BALLOON in which the two O’s are
side-by-side. For example, OOBALLN is one such arrangement, but BALLONO isn’t.
SOLUTION: 360 We treat the O’s as one giant letter (Call it Ω). Then, there are 6
letters, of which two are repeated (the L’s). Thus, our answer is 6!
= 360.
2!
3. Let a, b, c, and d be non-negative integers. How many ordered 4-tuples exist such that
a + b + c + d = 11?
SOLUTION: 364 Consider 11 balls lined up in a row. Now, drop three “walls” in the
middle of the balls to separate them into four groups (where one or more groups may be
empty). For example, if the walls went like this – o—oo—ooo—ooooo – then the ordered
quadruple would be (1, 2, 3, 5). Consider that what we are actually doing is arranging 14
14 · 13 · 12
14!
=
= 364.
things, 3 of which are walls. Thus, we compute 14
=
3
3! · 11!
3·2·1
Authors: Michael Curry - [email protected] – George Reuter - [email protected]
www.mathmeets.com
December 4, 2012
Calculators are not permitted on this contest.
The word “compute” calls for an exact answer in simplest form.
6 - Permutations and Combinations (No Probability)
1. Three identical red blocks and three identical blue blocks are arranged in a row of six
blocks. Compute the number of possible distinct arrangements.
2. The Tour de Citypolis is run in an 8-block by 4-block area. Each competitor must take
a different 12-block route to get from the southwest corner of Citypolis to the northeast
corner. Compute the maximum possible number of racers in the Tour.
3. Alice, Ben, Carol, David, and Ethan are seated on stage in six seats arranged in a row.
One seat remains empty. Compute the number of distinct arrangements in which at least
one person is seated between Alice and Carol but no seats between Alice and Carol are
empty.
December 2012
6 - Permutations and Combinations (No Probability)
1. Three identical red blocks and three identical blue blocks are arranged in a row of six
blocks. Compute the number of possible distinct arrangements.
SOLUTION: 20 We are arranging 6 items, so we have 6! = 720 ways to do this.
However, two sets of three identical items are used, so we divide out twice by 3! = 6, so our
answer is 720
= 20.
6·6
2. The Tour de Citypolis is run in an 8-block by 4-block area. Each competitor must take
a different 12-block route to get from the southwest corner of Citypolis to the northeast
corner. Compute the maximum possible number of racers in the Tour.
SOLUTION: 495 Suppose a racer starts at the southwest corner and must arrive at the
northeast corner of Citypolis. You’re really looking for the number of ways to arrange eight
“east” decisions and four “north” decisions, since those
are the only ways to get to the
12
desired corner. The number of ways to do this is 4 = 495.
3. Alice, Ben, Carol, David, and Ethan are seated on stage in six seats arranged in a row.
One seat remains empty. Compute the number of distinct arrangements in which at least
one person is seated between Alice and Carol but no seats between Alice and Carol are
empty.
SOLUTION: 240 In all cases, we (1) pick seats for Alice and Carol, (2) seat Alice and
Carol in these seats, (3) fill in the in-between seats, and (4) fill in the other seats, leaving
one empty.
Case 1 – 1 person in between – 4 · 2 · 3 · 3! = 144
Case 2 – 2 people in between – 3 · 2 · (3 · 2) · 2! = 72
Case 3 – 3 people in between – 2 · 2 · (3 · 2 · 1) · 1 = 24
This gives us a total of 144 + 72 + 24 = 240.
Authors: Michael Curry - [email protected] – George Reuter - [email protected]
www.mathmeets.com
December 13, 2011
Calculators are not permitted on this contest.
The word “compute” calls for an exact answer in simplest form.
4 - Permutations and Combinations (No Probability)
1. Compute the number of distinct arrangements of the letters of the word DECEMBER
that end in EEE.
n
2. Suppose that
denotes the number of ways to choose r things from a group of n
r
n
3
things without respect to order. If = 10, compute n.
n
2
3. The letters in the word N OT EBOOK are arranged so that no two vowels are adjacent
and no two consonants are adjacent. Compute the number of distinct arrangements of this
type. For example, N OT EBOKO and T ON OBOKE are two such arrangements.
December 2011
4 - Permutations and Combinations (No Probability)
1. Compute the number of distinct arrangements of the letters of the word DECEMBER that
end in EEE.
SOLUTION: 120 We ignore the three E’s and arrange the other five letters, so our answer is
5! = 120.
n
denotes the number of ways to choose r things from a group of n things
r
n
3
without respect to order. If = 10, compute n.
n
2
n(n − 1)(n − 2)
3·2·1
SOLUTION: 32 The equation may be rewritten as
= n−2
3 = 10, so n = 32.
n(n − 1)
2·1
2. Suppose that
3. The letters in the word N OT EBOOK are arranged so that no two vowels are adjacent and no
two consonants are adjacent. Compute the number of distinct arrangements of this type. For
example, N OT EBOKO and T ON OBOKE are two such arrangements.
SOLUTION: 192 Because there are 4 vowels and 4 consonants, we know the arrangement
must be vcvcvcvc or cvcvcvcv. There are 4! = 24 ways to place the vowels and 4! = 24 ways to
place the consonants. Thus, there are 24 · 24 = 576 ways to place the letters in the pattern
vcvcvcvc. So, you’d think the answer is just 576 · 2 = 1152. However, three of the letters are O’s,
so we divide by 3! = 6 to obtain 1152 ÷ 6 = 192.
DECEMBER 14, 2010
NO CALCULATORS ON THESE TOPICS
#4 PERMUTATIONS and
COMBINATIONS (no probability)
ANSWERS:
1.
608,400
2.
112
3.
30,240
1. A personal password for an internet brokerage account is to consist of a digit,
followed by two letter, followed by two digits. Find the number of personal
codes possible if the first digit cannot be zero and repetition is permitted.
2. Eight couples attend a benefit at which two prizes are given. In how many ways can two
names be randomly drawn so that the prizes are not awarded to the same couple?
3. Tom, Susan, Nancy, Rob, Yao, Patel, Lee, and Tracy are lining up for a picture. In how
many ways, from all possible arrangements, can this be accomplished so that Tom and Susan
are NOT next to each other?
DECEMBER 15, 2009
NO CALCULATORS ON THESE TOPICS
#4 PERMUTATIONS and
COMBINATIONS (no probability)
1.
ANSWERS:
1.
262
2.
5250
3.
20
P 10 P5
+
=?
3!
5!
5 3
2. Euclid High has 20 students on their math team. 6 are seniors, 7 are juniors,
5 are sophomores, and the rest are freshmen. How many ways can their A Team
be selected if it consists of 4 seniors, 4 juniors, and 2 sophomores?
3. Each person at a dinner table shakes hands with everyone excect the two people on either
side. How many handshakes will there be among 8 diners?
DECEMBER 9, 2008
NO CALCULATORS ON THESE TOPICS
#4
PERMUTATIONS and
COMBINATIONS (no probability)
ANSWERS:
1.
72
2.
6
3.
50
1. A musical piece has 2 saxophone parts, 3 trumpet parts, and 3 trombone parts. If a band has
2 saxophonists, 3 trumpeters, and 3 trombonists, in how many ways can these parts be
assigned to the players?
2. How many different 4-digit numbers can be formed by arranging the digits 2,0,0,8?
3. A chemist has nine samples of a solution, of which four are type A and five are type B. If the
chemist chooses three of the solutions at random, determine how many ways the chemist can
have at most one type A solution.
)yION]ROIECOUNTY
N[AT]I{ T,]EAGIUJE
DECEMBER9,2OO8
NO CALCALATORS ON THESE TOPICS
#4
PERMUTATIONS and
COMBINATIONS (noprobability)
ANSWERS:
1.
72
2.6
50
3.
parts,3 trumpetparts,and3 tromboneparts. If a bandhas
1. A musicalpiecehas2 saxophone
2 saxophonists,
3 trumpeters,and 3 trombonists,in how manywayscanthesepartsbe
assignedto the players?
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2. How manydifferent4-digit numberscanbe formedby anangingthe digits 2,0,0,8?
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3. A chemisthasnine samplesof a solution,of which foru areqngA andfivrtare type.B. If the
chemistchoosesthreeof the solutionsat random,determinehow manywaysthe chemistcan-
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DECEMBERi 1,2OO7
NO CALCULATORSOIYTHIS TOPIC
#4
PERMUTATIONS
and
ANSWERS:
COMBINATIONS 1noprobability)
1
5040
2
720
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DECEMBER
12,2006
]YO CALCULATORSON THIS TOPIC
#4
PERMUTATIONSand
COMBINATIONS
ANSWERS:
1noprobability)
72
144
l. How many gamescan be playedin a 9-teamsportsleagueif eachteam plays
all other teamstwice?
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tlrelettersarenot repeated,
but mustbeginwith DE. Findp + q.
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If 3 of thesecardsare chosenat random,in how many ways will there exist apair?
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