Combinatorics

Reference for the ICTM Regional Math Contest

Foreword Many thanks to Richard Rhoad and Robert Whipple. They created a document in 1983 which laid the foundation for this document.

The study of combinatorics and probability is one of the most complex studies in mathematics. Advanced collegiate courses exist in both the study of combinatorics and the study of probability. This document provides a preview of what each of those studies entails.

A few introductory notes: • In Chapter 1, all numbers presented will be a subset of whole numbers.• Many problems have more than one means to obtain an answer. While one solution is

provided, any mathematically sound, well-explained solution will be acceptable during the actual competition.

Before delving into these topics, some background information is required: • All ‘coins’ are fair with a heads on one side and a tails on the other, and the result(s) of a

“coin-flip” or “coin-flips” is/are either heads or tails (with equal probability) on each flip.• Become familiar with a die (singular) and dice (plural). All references to ‘die’ or ‘dice’

refer to a six-sided die with equiprobable weights unless otherwise stated.• Become familiar with a deck of cards. References to a ‘bridge deck’ or ‘standard deck’

refer to a 52-card deck with a 2 through Ace (rank) each of clubs, diamonds, hearts andspades (suits). Note: The two jokers found in some decks of cards will not be consideredin any problem.

• The word ‘vowel’ refers to a, e, i, o, and u. The letter y in this reference is considered aconsonant.

Table of Contents

Chapter 1 – Combinatorics 1.1 – The Fundamental Principles A – Fundamental Principle of Multiplication………………………………………… 1 B – Fundamental Principle of Addition……………………………………………….1 Problem Set……………………………………………………………………………3 1.2 – Overlapping Possibilities A – Principle of Inclusion-Exclusion – Two Sets………………………………….... .5 B – Principle of Inclusion-Exclusion – Three Sets…………………………………....6 Problem Set……………………………………………………………………………9 1.3 – Factorials and Permutations A – Factorials………………………………………………………………………...11 B – Permutations.…………………………………………………………………….11 Problem Set…………………………………………………………………………..14 1.4 – More on Permutations A – Permutations with Restriction…………………………………………………...15 B – The Backdoor Method………………………………………………………….. 16 C – Permutation with Repetition (Ordered Partitions) ……………………………... 17 D – Circular Permutations…………………………………………………………... 18 E – Permutations Involving Bracelets………………………………………………. 19 Problem Set…………………………………………………………………………..21 1.5 – Combinations.………………………………………………………………………. .24 Problem Set…………………………………………………………………………..27 1.6 – Binomial and Multinomial Theorems A – Summation Symbol…………………………………………………………….. 30 B – The Binomial Theorem…………………………………………………………. 31 C – The Multinomial Theorem……………………………………………………… 32 Problem Set…………………………………………………………………………..37 1.7 – Combinatorial Equations A – Expressions and Equations with Factorials…………………………………….. 39 B – Combinatorial Equations………………………………………………….……. 41 Problem Set…………………………………………………………………………..45 1.8 – Combinatorics……………………………………………………………………….. 48 Solutions to Problem Sets…………………………………………………………………….. 60

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1.1 – Two Fundamental Principles 1.1(A) Fundamental Principle of Multiplication EX 1 – A license plate must contain two letters followed by four digits with the first digit being

non-zero. Determine the number of different license plates possible. Solution: 26 26 9 10 10 10 6,084,000

EX 2 – Determine the number of odd natural numbers which contain two digits.

Solution: You must have a ten’s digit AND a unit’s digit. The ten’s digit can be 1, 2, 3, 4, 5, 6, 7, 8, or 9. The unit’s digit can be 1, 3, 5, 7, or 9. AND means MULTIPLY, so 9 5 45 .

1.1 (B) Fundamental Principle of Addition EX 3 – Maggie places a $1 bill and a $5 bill on her desk. Determine how many distinct positive

amounts of money Zeke can create from the bills on Maggie’s desk.

Solution: If you use 1 bill, there are 2 possibilities ($1 or $5). If you use 2 bills, there is 1 possibility ($6). 2 + 1 = 3.

If a procedure can be performed in 1n different ways, and if, following this first procedure, a second procedure can be performed in 2n different ways, etc.; then the number of ways ALL these procedures can be performed in the order indicated is the PRODUCT 1 2 3n n n ...

If an answer requires the first occurrence AND the second occurrence, etc., you MULTIPLY the possibilities for each occurrence.

If a procedure can be performed in 1n different ways, and if, following this first procedure, a second procedure can be performed in 2n different ways, etc.; then the number of different ways in which the first procedure OR the second procedure, etc., can be performed is the SUM 1 2 3n n n ...

If an answer requires the first occurrence OR the second occurrence, etc., you ADD the possibilities for each occurrence.

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EX 4 – Anne’s cable television package only allows her to see channels 2, 5, 7, 9, and 11. Each of these channels runs three different 30-minute programs (one from 8:00-8:30, one from 8:30-9:00, and one from 9:00-9:30). From 8:00 until 9:30 each Tuesday night, Anne watches three one-half hour programs. Determine the number of ways she can spend this part of the evening.

Solution: At 8:00 Anne has her choice of 5 programs. At 8:30 Anne has her choice of 5

programs. At 9:00, Anne has her choice of 5 programs. The number of ways ALL these procedures can be performed is the PRODUCT 5 5 5 125 .

EX 5 – In taking one of Doc’s true-false tests, Laura answers each question. If there were 10

questions on the test, a) Determine the number of distinct ways Laura could have answered the test.

b) If Laura could also leave any question blank, determine the number of distinct ways Laura could have answered the test.

Solution: Using the Fundamental Principle of Multiplication, a) Since each question could be answered in 2 ways, there are 102 possibilities. b) Since each question could be answered in 3 ways, there are 103 possibilities.

EX 6 – One day in a class of 25 geometry students, the teacher selected 3 students at random.

He awarded the first student 3 extra credit points; the second, 2; and the third 1. Determine the number of ways this wonderful surprise could have taken place.

Solution: There are 25 possibilities for the first surprise. There would be 24 possibilities for

the second surprise, AND 23 possibilities for the third. Using the Fundamental Principle of Multiplication, there are 25 24 23 13,800 possibilities.

EX 7 – Determine how many numbers of at most three digits can be formed from the digits 1, 2,

3, 4, or 5 if all digits in a particular number must be different.

Solution: One Digit Numbers: 5 Two digit numbers require a first AND second digit: 5 4 20 Three digit numbers require a first, second AND third digit: 5 4 3 60 Since there could be a one digit, two digit, OR three digit number:

5 20 60 85

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1.1 – Two Fundamental Principles – Problem Set 1. From a team of 12 basketball players, determine the number of ways a starting lineup of 5

players can be announced. 2. Determine how many three-digit numbers less than 500 can be formed from the integers 3,

4, 5, and 9 if: a) the three digits must be different. b) the three digits need not be different. 3. In her closet, Jenny has 13 sweaters, 17 skirts, and 21 pairs of shoes. Determine how many

different outfits Jenny can wear if an outfit consists of one sweater, one skirt, and one pair of shoes.

4. At a hamburger joint, one can order a hamburger rare, medium, or well-done. It is served

on either a bun or dark rye bread. A single topping may be added from: onion, mustard, relish, pickle, ketchup. Determine the number of different types of hamburgers are possible.

5. At Mathlete University, a student ID code consists of two letters followed by four digits.

Determine the number of ID codes are possible: a) if repetitions are allowed. b) if repetitions are not allowed. c) if repetitions are not allowed, and the first digit must be non-zero. 6. Determine the number of ways 8 different novels can be arranged on a bookshelf if your

favorite novel must be on the far left. 7. A national mathematics exam consists of 30 multiple choice questions. Each question has

five different choices, or a student may leave any question blank. Determine the number of ways Malcolm could complete the exam.

8. Determine how many numbers of at least 4 digits and no more than 7 digits can be formed

from the integers 2, 3, 4, 5, 6, 7, or 8 if repetition of digits is allowed. 9. From 7 different candy bars, Mr. Hershey gives one to Jerry, one to Nancy, and one to

Adam. Determine the number of different ways this can be done. 10. Determine how many license plates of 3 symbols (letters and digits) can be made using at

least one letter in each. 11. Determine how many odd numbers between 10 and 1000 start and end with the same digit. 12. A license plate consists of any two different letters followed by four different digits with

the first being non-zero. Determine the number of possible license plates.

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13. DNA molecules include the base units adenine, thymine, cytosine, and guanine (A, T, C, and G). The sequence of base units along a strand of DNA encodes genetic information. Determine how many different sequences the 4 base units A, T, C, and G can be arranged along a short single strand of DNA that includes 8 base units.

14. You have completely forgotten the combination to your locker, which has three numbers in

order (e.g., 31-2-15 or 30-30-2). Each number is between 0 and 35 (inclusive). If you can test one combination every 15 seconds, determine how many minutes it will take you to test all possible combinations.

15. There are 5 ways to get from A to B, 4 from A to C, 6 from B to C, 3 from B to A, 4 from

C to A, and 2 from C to B. a) Determine how many ways can Wendy go directly from A to B to C or B to A to C. b) Determine how many ways she can go from any letter to a second and then to the third. c) Determine how many ways she can go from any letter to a second and then either to the

third or back to her starting letter. 16. Let A = {1, 2, 3}, B = {4, 5, 6, 7} and C = (8, 9, 10, 11, 12}. John can choose 1 element

from each of the three sets or 1 element from any two of the three sets. Determine the number of different possible combinations John can choose.

17. At Mathlete High School, every senior must take 5 core classes during the first five periods,

lunch during sixth period, study hall during seventh period, PE during eighth period, and an elective during the ninth period. Seniors have 6 core classes and 4 electives available to them.

a) Determine the number of different ways a schedule can be arranged. b) Explain how your answer to part a would change if lunch was offered during fourth,

fifth, or sixth period. 18. Ray’s row consists of 3 students, Laura’s row has 5 students, David’s row has 6 students,

and Dianne’s row has 4 students. A committee of three students is formed, with one student from each of three of the four rows. Determine the number of possible committees.

19. A football stadium has 16 gates. Determine the number of ways Jim can enter one and

leave by a) a different gate. b) any gate. 20. In selecting an Ace, King, and Queen from a standard deck of cards, determine the number

of ways they can be chosen if: a) they must be of different suits. b) they may or may not be of different suits. c) they must be of the same suit.

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1.2 – Overlapping Possibilities Consider this problem: In how many ways can you draw either an Ace or a Heart from a standard deck of cards?

Issue: There are 4 aces and 13 hearts, so the answer should be 17, but it is not. What about the Ace of Hearts? That has been counted twice (once as an Ace and once as a Heart). See below for solution.

In some cases, sets may overlap, like the Ace of Hearts, which is counted in the set of Aces and set of Hearts. How do we compute n A B when there are common elements between the two sets?

1.2(A) – Principle of Inclusion-Exclusion – Two Sets

Back to our original problem: In how many ways can you draw either an Ace or a Heart from a standard deck of 52 cards?

Solution: Let A be the set of aces and B be the set of hearts.

n A 4, n B 13, n A B 1

n A B 4 13 1 16

EX 1 – Two intersecting circles are drawn on the floor. Twenty students are standing in circle

A. Fifteen students are standing in circle B. If there are 26 students in the class, and every student is standing in at least one circle, determine the number of students standing in A B.

Solution: n A B n A n B n A B 26 20 15 n A B n A B 9

This problem can also be symbolized with a Venn diagram, as shown below.

Cardinal Number of a Set The cardinal number of set A, denoted n(A), is the number of elements in A.

Principle of Inclusion-Exclusion – Two Sets

n A B n A n B n A B

If n A B 0 , sets A and B are mutually exclusive sets.

A B 11 9 6

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1.2(B) - Principle of Inclusion-Exclusion – Three Sets EX 2 – Out of 100 sophomores, 73 are taking Geometry, 49 are taking French, 61 are taking

Chemistry, 38 are taking French and Geometry, 45 are taking Geometry and Chemistry, 24 are taking French and Chemistry, and 10 students are not taking any of the three. Determine the number of students who are taking all three courses.

Solution: Using G for Geometry, F for French, and C for chemistry, n G 73; n F 49; n C 61; n G F 38; n G C 45; n F C 24

Since 10 people are taking any one of the three, there are 90 students taking at least one of the courses: n G F C 90

90 73 49 61 38 45 24 n G F C

n G F C 14

Alternative Solution: This can also be solved using a Venn Diagram, starting from the inside and working out.

Principle of Inclusion-Exclusion – Three Sets

n A B C n A n B n C n A B n A C n B C n A B C

G F

C 10

x

G F

C 10

x

38 – x

24 – x 45 – x

G F

C 10

x

38 – x

24 – x 45 – x

x – 10 x – 13

x – 8

The sum of all 8 expressions shown is x + 86 x 86 100 x 14

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EX 3 – Out of a class of 25 students, 20 are taking science and 17 are taking history. Assuming each student takes at least one, determine the number of students who are taking both.

Solution: n S H n S n H n S H

25 20 17 n S H

n S H 12

EX 4 – From the set {1, 2, 3, …, 50}, determine how many numbers are divisible by either 2

or 3.

Solution: n(divisible by 2) = 25, n(divisible by 3) = 16 n(divisible by 2 and 3) = n(divisible by 6 ) = 8 25 16 8 33

EX 5 – There are 80 juniors at Mathlete High School. 35 juniors study only French, 9 juniors

study both French and German, and 2 juniors study neither French nor German. Determine the number of juniors who study only German.

Solution: n F G 80 2 78, n F 35 9 44, n G x 9, n F G 9 n F G n F n G n F G 78 44 x 9 9 x 34

S H

20 – x x 17 – x

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EX 6 – Sporty asked 80 people which sports they enjoyed from football, hockey and baseball. 4 liked only football, 3 liked only baseball, and 5 liked only hockey. 45 liked football and hockey, 48 liked football and baseball, and 36 liked hockey and baseball. In addition, 31 liked all three. Determine the number of people who did not like any of the three sports.

Solution: n F H B 80 x, n F 4 14 31 17 66, n H 5 14 5 31 55

n B 3 5 17 31 56, n F H 14 31 45, n F B 17 31 48,

n H B 5 31 36, n F H B 31

n F H B n F n H n B n F H n F B n H B

n F H B

80 x 66 55 56 45 48 36 31 x 1 EX 7 – 25 female IHSA athletes were surveyed regarding the sport(s) in which they participate:

Fall Tennis (T), Winter Basketball (B), and Spring Softball (S). Given the Venn Diagram below, determine the number of athletes who participated in all three sports.

Solution: 2 3 6 1 x 8 0 1 25 x 4

S T

B 0

x

3

8 1

6 2

1

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1.2 – Overlapping Possibilities – Problem Set 1. At a dinner party, guests choose from ice cream and cake for dessert. 60 chose ice cream,

43 chose cake, and 9 chose both ice cream and cake. If 107 people attended the party, determine the number of guests who opted for no dessert.

2. In a certain class there are 21 students who like Minecraft and 17 students who like World

of Warcraft. If 12 of the students like both games, and all students like at least one of the games, determine the number of students in the class.

3. A poll taken at the county fair involved 18 participants. When asked what flavor ice cream

they like, 11 liked vanilla, 10 liked chocolate, 8 liked cookies and cream, 3 liked vanilla and chocolate, 5 liked vanilla and cookies and cream, and 5 liked cookies and cream and chocolate. All 18 participants responded at least one of the three flavors above.

a) Determine the number of participants who liked all three. b) Determine the number of participants who liked only cookies and cream. 4. From the set {1, 2, 3, …, 200}, determine how many numbers are divisible by either 3 or 5. 5. The local fitness center runs two classes: spinning and circuits. On Saturday, 100 people

visited the fitness center. 18 people attended the spinning class. 10 people attended both classes. 56 people did not attend either class. Determine the number of people who attended exactly one class.

6. The Venn diagram below describes the 150 cars currently in the parking garage, where R is

a red colored car and A is a car assembled in America. Determine the number of cars assembled in America which are currently in the parking

garage.

A R

2x + 5 x - 7 21

68

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7. A survey was taken of the leisure time activities of 90 students. 60 students play video games, 60 students read, 70 students go to the movies, 26 students do all three, 20 students play video games and go to the movies only, 18 students read and go to the movies only, and 10 students read and play video games only. If all students participate in at least one of the three activities, a) determine the number of students that only play video games. b) determine the number of students that only go to the movies.

8. A total of 135 students were surveyed about their favorite science classes. Results showed

that 63 like Physics; 61 like Chemistry; 53 like Biology; 28 like Chemistry and Biology; 31 like Physics and Chemistry; 22 like Physics and Biology, and 15 like all three courses. a) Determine the number of students who do not like ANY of the science courses. b) Determine the number of students who liked only Physics.

9. In a class of 24 students, 12 students play piano, 13 students play guitar, and 4 students

play neither instrument. Determine the number of students who play both piano and guitar. 10. The opening of a new gaming store was publicized through direct mail, the internet, and

television. A total of 119,000 customers received direct mail; 131,000 customers learned about the opening through the internet; 140,000 customers got to know about it through television; 25,000 customers got to know about it through direct mail and internet, but not television; 19,000 customers were aware of the opening through direct mail and television, but not internet; 27,000 customers were aware of the opening through internet and television, but not through direct mail; and 16,000 customers were aware of the opening from all three sources. a) Determine the number of customers who were aware of the opening only through

television. b) Determine the number of customers who were aware of the opening through at least

two of the three modes of advertisement used. 11. A number can be called “prime-looking” if it is a composite number (has factors other than

1 and itself) that is not divisible by 2, 3, or 5. There are 168 prime numbers less than 1000. Determine the number of “prime-looking” numbers less than 1000.

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1.3 – Factorials and Permutations 1.3(A) Factorials

The product of the positive integers from 1 to n inclusive is denoted by n! (read “n factorial”).

EX 1 – Expand and simplify 6!. EX 2 – Expand and simplify 5!2!0!

.

Solution: 6 5 4 3 2 1 720 Solution:

5 4 3 2 1

602 1 1

1.3(B) Permutations

EX 3 – List the permutations of the letters A, B, and C if all three are used, and no letter is

repeated.

Solution: One permutation is ABC. A second permutation is ACB. There are a total of six permutations, the other four being BAC, BCA, CAB, and CBA.

EX 4 – List the possible permutations of the letters W, X, Y, and Z when taken two at a time if no letter is repeated.

Solution: WX, WY, WZ, XY, XZ, YZ, as well as all of them in reverse order: XW, YW, ZW, YX, ZX, ZY. There are 12 permutations possible.

Definitions of Factorial

n! n n 1 n 2 ... 3 2 10! 1

The key to the definition of a permutation is the word ORDER.

Become comfortable expanding and simplifying expressions with factorials by hand. This will be a necessary skill in Section 1.7.

Definition of Permutation

Each different arrangement or ORDERED set of objects is called a PERMUTATION of those objects.

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In the study of combinatorics, we are more interested in the “number” of permutations than the permutations themselves.

EX 5 – Expand and simplify P(7, 5).

Solution: P 7, 5 7 7 1 7 2 ... 7 5 1 7 6 5 4 3 2520 or

7 6 5 4 3 2 17!P 7, 5 2520

7 5 ! 2 1

EX 6 – (Refer to EX 3) Determine the number of permutations of the letters A, B, and C if all

three are used, and no letter is repeated.

Solution: There are three different letters taken one at a time:

3 2 13!P 3, 3 6

3 3 ! 1

EX 7 – (Refer to EX 4) Determine the number of permutations of the letters W, X, Y, and Z when taken two at a time if no letter is repeated.

Solution: There are four different letters taken two at a time:

4 3 2 14!P 4, 2 12

4 2 ! 2 1

Permutation Theorem 1

The number of permutations of “n” different objects taken “r” at a time is denoted by P(n, r).

P n, r n n 1 n 2 ... n r 1 or

n!P n, r

n r !

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EX 8 – (Refer to Section 1.1, Example 6) One day in a class of 25 geometry students, the teacher selected 3 students at random. He awarded the first student 3 extra credit points; the second, 2; and the third 1. Determine the number of ways this wonderful surprise could have taken place.

Solution: Think of the problem as arranging 25 different students 3 at a time.

25 24 23 22 21 ... 125! 25!P 25, 3 13,800

25 3 ! 22! 22 21 ... 1

EX 9 – If no digit can be repeated, determine the number of ways one can make a 2-digit

number if one chooses from the set {1, 2, 3, 4, 5}.

Solution: Think of the problem as arranging 5 different digits 2 at a time.

5 4 3 2 15! 5!P 5, 2 20

5 2 ! 3! 3 2 1

EX 10 – Determine the number of 5-letter arrangements that can be formed from the letters in

the word SINGAPORE.

Solution: Think of the problem as arranging 9 different letters 5 at a time.

9 8 7 6 5 4 3 2 19! 9!P 9, 5 15,120

9 5 ! 4! 4 3 2 1

Note the emphasis on the word ‘different’. In the next section, we will examine permutations where some objects appear more than once.

Note: There are three notations for the permutation.

n r

nP n, r P P

r

For the purposes of this document, we will use P(n, r).

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1.3 – Factorials and Permutations – Problem Set 1. Expand and simplify each expression below.

a) 4! b) P(6, 3) c) 9!7!

d) 8!6!2!

e)

n!n 3 !

f) P(8, 2)

g)

2n!n 1 ! n 2 !

h) P(4, 4) i) 7!5!0!

2. Student Council has four officials: president, vice-president, secretary, and treasurer.

Assuming a member cannot hold more than one office, determine the number of ways officials can be elected if the Council has:

a) 12 members b) 16 members 3. Determine the number of ways the starting lineup of 5 players can be announced for a

basketball team with 11 players. 4. Determine the number of three-digit numbers which exist using the digits 1, 2, 3, 4, 5, and

6 if no digit can be used more than once for any single number. 5. A shelf will only hold an even number of books. Given that 9 different books are

available, determine the number of different arrangements of at least one book that can be made on the shelf.

6. 5 boys and 8 girls are available to perform at the school play. The play has 3 different

male roles and 4 different female roles. Determine the number of possible cast lists if the male roles are listed before the female roles.

7. Given the letters U, V, W, X, Y, and Z, arrangements of 3, 4, or 5 letters are created.

Determine the number of possible arrangements if no letter can be repeated in any particular arrangement.

8. A gym locker combination consists of three different numbers from 0 to 9 inclusive.

Determine the number of possible combinations.

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1.4 – More on Permutations 1.4 (A) Permutation with Restriction Consider the problem below from Section 1.3 (Example 5):

If no digit can be repeated, determine the number of ways one can make a 2-digit number if one chooses from the set {1, 2, 3, 4, 5}.

Now consider this updated problem:

If no digit can be repeated, determine the number of ways one can make a 2 digit even number if one chooses from the set {1, 2, 3, 4, 5}

Solution: The fact that the number must be even restricts the possibilities for the final

digit…must be 2 or 4. The final digit has two possibilities (choose 1), leaving four possibilities for the first digit. 4 2 8

EX 1 – Determine the number of automobile license plates that can be made if each plate

contains 3 different letters followed by 3 different digits with the first digit being non-zero.

Solution: Three different letters: P(26, 3) = 15,600. Since the first digit must be non-zero,

there are only 9 possibilities. Once that digit is selected, it can’t be used, but 0 could be the second digit. 15600 9 9 8 10,108,800

EX 2 – Determine the number of ways 6 students can sit in a car if only 3 of them have licenses.

Solution: Only 3 can sit in the driver’s seat. Once that is determined, the remaining five students can sit anywhere. 3 5 4 3 2 1 360

Permutation with Restriction

When a permutation contains some sort of restriction, process the restriction first. Complete the computation using an appropriate combination of the Multiplication Principle (1.1A), Addition Principle (1.1B), and Permutation Theorem 1 (1.3B).

16

1.4 (B) The Backdoor Method Consider the problem below:

A 6-digit number is created from the set {1, 2, 3}. Digits may be repeated. Determine the number of 6-digits numbers which contain at least one 3.

Solution: Overall, there are 63 729 6-digit numbers that can be created. How many of

them contain at least one 3? Take the 729 and subtract the 6-digit numbers that contain no 3’s: 6 63 2 729 64 665

EX 3 – The serial numbers of appliances contain 3 letters followed by 4 digits. Determine the

number of serial numbers which contain at least one 0.

Solution: Overall, there are 26 26 26 10 10 10 10 175,760,000 . Subtract those that have no 0’s: 26 26 26 9 9 9 9 115,316,136

175,760,000 115,316,136 60,443,864 EX 4 – From the letters of the word DECIMAL, arrangements of 3 different letters are formed.

Determine the number of permutations which contain at least one vowel.

Solution: Overall, there are P 7, 3 210 arrangements. Subtract those that have no vowels: P 4, 3 24 . 210 – 24 = 186

Backdoor Method

If a permutation with restriction has many options, consider computing the grand total number of possibilities (ignoring restriction) and subtract those possibilities which do not meet the restriction.

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1.4 (C) Permutation with Repetition (Ordered Partitions) Consider the problem below:

Determine the number of permutations using all three letters of the word BOB.

Issue: If we thought of the word as 1 2B OB , a list of all possible permutations would be: B1OB2, B2OB1, B1B2O, B2B1O, OB1B2, OB2B1. Notice that the list includes both

1 2B OB and 2 1B OB , which are essentially the same arrangement. We can compute the number of permutations using Permutation Theorem 2 – Repetition.

.

Solution: The number of permutations using all three letters of the word BOB is 3! 32!1!

.

EX 5 – Determine the number of permutations using all eight letters of the word GEOMETRY

Solution: There are eight total letters: two E’s, one G, one O, one M, one T, one R, and one

Y. Therefore, the number of possible permutations is 8! 201602!1!1!1!1!1!1!

Note: While the 1! is not necessary since 1! = 1, it is presented here to show that all 8 letters have been accounted for.

EX 6 – Determine the number of ways 12 different candy bars can be distributed to Marcia,

Jennifer, and Felicia if Marcia gets 5, Jennifer get 4, and Felicia gets 3. Solution: If we think of permuting the people, we obtain the ‘word’ MMMMMJJJJFFF.

The number of ways to permute these letters is 12! 27,7205!4!3!

.

Permutation Theorem 2 - Repetition

The number of permutations of “n” objects of which “p” are alike, “q” are alike, “r” are alike, etc., is:

n!p!q!r!...

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EX 7 – Determine the number of ways 3 identical emeralds, 2 identical diamonds, and 2 identical opals can be arranged in a row in a display case.

Solution: In Section 1.3, the emphasis was on the word ‘different’. Here, the emeralds,

diamonds, and opals are ‘identical’, implying the use of Permutation Theorem 2 – Repetition.

7! 210

3!2!2!

1.4 (D) Circular Permutations Consider the problem below:

Determine the number of ways 3 people can be seated at a round table.

Solution: Look at these three diagrams:

If you are person A, B is seated to your left, and C is seated to your right in all three configurations. In a circle, for any particular order such as ABC, we consider BCA and CAB as the same clockwise order. Therefore, the answer is 3!3

which is the same as 2!.

EX 8 – Determine the number of ways 8 people can be seated in a circle.

Solution: 8 1 ! 7! 5040

Circular Permutations

The number of permutations of “n” objects arranged in a circle with no restrictions is n 1 !

A B

C

C A

B

B C

A

19

EX 9 – Twelve different jewels are to be placed in a circle. Determine the number of possible arrangements of the jewels.

Solution: 12 1 ! 39,916,800

1.4 (E) Permutations Involving Bracelets When considering bracelets, another oddity occurs. Imagine you are looking at the bracelet below:

Now turn the bracelet over, and you see: On one side, the clockwise order is ABC, but the same bracelet appears in the clockwise order of BAC when turned over. Normally, we consider that we cannot distinguish one side of a bracelet from the other. Thus, we must divide our expected circle permutation answer by 2. EX 10 – Determine the number of ways 6 different colored beads can be arranged on a bracelet.

Solution: 6 1 ! 5! 60

2 2

Bracelet Permutations

The number of permutations of “n” objects arranged on a

bracelet is n 1 !.

2

A B

C

Front Side

B A

C

Back Side

20

EX 11 – JoAnn has a birthday bracelet. Beginning with her first birthday, her mom has added another trinket to her bracelet on JoAnn’s bracelet. Determine the number of ways her trinkets can be arranged on her bracelet on her 8th birthday.

Solution: 8 1 ! 7! 2520

2 2

EX 12 – Determine the number of ways 2 tennis players, 5 basketball players, and 9 baseball

players can be seated in a row for the pep rally if those in the same sport sit together. Solution: Group the players by sport 2!5!9! . The 3 sports can be arranged in 3! ways. 2!5!9!3! 522,547,200

EX 13 – Consider the 6 people: Abigail, Ben, Carlos, Diane, Emily, and Frank. a) Determine the number of ways in which all six can sit in a row.

b) Determine the number of ways in which all six can sit in a row with Abigail sitting next to Ben.

c) Determine the numbers of ways in which all six can sit in a row if the sexes must alternate.

Solution: a) Using Permutation Theorem 1, 6! 720

b) Since two people must be a single unit, there are now 5 units instead of 6. Abigail and Ben can sit together in 2! ways. 5!2! 240

c) The first seat can be filled by any of 6 people. Since the sexes must alternate, the second seat could then be filled by any of the 3 people of the opposite sex, and so on. 6 3 2 2 1 1 72

EX 14 – In computer programming, a byte is defined as an eight digit number with all 0’s or 1’s. Determine the number of bytes which contain at least two 1’s.

Solution: Overall, there are 82 256 bytes possible. Subtract those that contain no 1’s. (1 possibility….00000000)

Subtract those than contain exactly one 1. (8 possibilities… 8! 87!1!

)

256 – 1 – 8 =247

21

1.4 – More on Permutations – Problem Set 1. Determine the number of ways 15 basketball players can be placed into a first team, second

team, and third team of five players each. 2. Determine the number of ways 5 boys and 4 girls be seated in a row containing 9 seats if: a) they may sit anywhere. b) the sexes must alternate. c) the sexes must be together. 3. Determine the number of 3-letter codes that can be formed if at least one of the letters is to

be chosen from the vowels A, E, I, O, and U. 4. Determine the number of ways 7 people can sit: a) in a row if 3 people insist on sitting together. b) around a circular table if 2 people insist on sitting together. 5. Determine the number of ways 6 friends can be seated in a car with 6 seats if only 3 have

licenses. 6. Determine the number of ways all the letters in the word ALGEBRA can be arranged. 7. Determine the number of ways 5 distinct keys can be arranged on a key ring. 8. There are 5 juniors and 5 seniors on the Student Council. Three are selected and arranged

in a line for the yearbook picture. Determine the number of arrangements in which at least 1 senior is in the picture.

9. On student recognition night, Mathlete High School will present awards to 4 seniors and 3

juniors. Determine the number of different orders the awards can be presented if: a) awards can be presented in any order. b) awards are presented to juniors before awards are presented to seniors. c) the first and last awards are presented to juniors. d) the first and last awards are presented to seniors.

22

10. Each member of a 9-member committee must be assigned to exactly one of three subcommittees: the executive subcommittee, the finance subcommittee, or the rules subcommittee. If these subcommittees are to contain 3, 4, and 2 members, respectively, determine the number of different subcommittee appointments that can be made.

11. A license plate must contain a sequence of 2 letters followed by 4 digits, or 3 letters

followed by 3 digits. Determine the number of license plates that contain the letter Q at least once.

12. Determine the number of ways 8 people can be arranged in a circle. 13. Five books (A, B, C, D, and E) are arranged on a shelf such that B is next to E. Determine

the number of arrangements that are possible. 14. To open the final door in an Escape Game, the group must enter a code of any two letters

followed by any three digits. Jamie knows the owner, who told her that the digits 0 and 8 are not part of the final code. Determine the number of possible codes Jamie must enter to open the final door.

15. In a certain district of a town there are ten streets running north-south, and five running

east-west. All streets/blocks are of equal length. Determine the number of ways a person can walk from the southwest corner of the district to the northeast corner, always taking the shortest course.

16. Determine how many numbers between 2000 and 5000 can be made from the digits 1, 2, 4,

5, 7, and 8 if each digit is used only once. 17. Grandma has sent a charm to Leona for every birthday beginning with her 10th birthday.

Leona adds them to her bracelet each year. Determine the number of ways Leona can arrange her charms after she receives the charm for her 16th birthday.

18. Determine the number of positive integers less than 1000 that can be formed using the

digits 1, 2, 3, 4, and 5, where no digit may be repeated, and the digit 2 must appear at least once.

23

19. Seven paintings are hung on a wall for an art show. Three of the paintings are created by the same artist. a) Determine the number of arrangements in which the artist’s 3 paintings are hung next to

each other. b) Determine the number of arrangements in which one of the artist’s 3 paintings is hung

at the beginning of the row, but neither of the other 2 is hung at the end of the row. 20. A dinner host and his four guests are about to sit down for dinner around a table. a) Determine the number of possible seating arrangements.

b) The host sits at the table first. Determine the number of possible seating arrangements for the remaining guests.

21. John just graduated from college and has a new job as a teacher. He has a building key for

his apartment building, key to his apartment, key to his school building, key to his classroom, and the FOB for his car. Determine the number of ways these can be arranged on his key ring.

22. Determine the number of ways 5 men and 5 women can be seated at a circular banquet

table if they must alternate the men and women. 23. If there are 4 chocolate chip, 2 oatmeal, and 2 double chocolate cookies in a box, determine

how many different orders are possible to eat all the cookies. 24. Determine the number of ways 5 similar bicycles and 4 similar fishing poles can be

distributed among 9 people, assuming each is to receive one item.

24

1.5 – Combinations Permutations are used when order is important. Combinations are used when order does not matter.

EX 1 – Expand and simplify C 8, 5 .

Solution: 8 7 6 5 4 3 2 18! 8!C 8, 5 56

8 5 !5! 3 !5! 3 2 1 5 4 3 2 1

EX 2 – Expand and simplify C(6, 2) and C(6, 4).

Solution: 6 5 4 3 2 16! 6!C 6, 2 15

6 2 !2! 4!2! 4 3 2 1 2 1

6 5 4 3 2 16! 6!C 6, 4 15

6 4 !4! 2!4! 2 1 4 3 2 1

Combination Theorem

The number of combinations of “n” objects taken “r” at a time is denoted by C(n, r).

n!C n, r

n r !r!

or

n n 1 n 2 ... n r 1C n, r

r r 1 r 2 ...1

Other notations for a combination include n r

nand C .

r

Since n!C n, r

n r !r!

, it can be seen that:

C n, r C n, n r

25

EX 3 – From 10 sophomores and 12 juniors, determine how many 5 person committees can be formed if each committee must consist of 2 sophomores and 3 juniors.

Solution: We are NOT interested in the ORDER of the students, but merely who they are.

We use COMBINATIONS and the Fundamental Principle of Multiplication.

10 9 12 11 1010! 12!C 10, 2 C 12, 3 99008!2! 9!3! 2 1 3 2 1

EX 4 – Given a standard 52-card deck, determine the number of ways a person can be dealt 5

cards of which 4 cards are from one suit and 1 card is from another suit.

Solution: Consider 3 factors: 1) The suit can be any of 4. 2) In any particular suit, we want C(13, 4). 3) The remaining card is C(39, 1). Since we need the first factor, and second factor, and third factor, the answer is

13 12 11 1013! 39! 394C 13, 4 C 39,1 4 4 111,5409!4! 38!1! 4 3 2 1 1

EX 5 – Determine the number of different positive sums of money that can be made from a

penny, a nickel, a dime, and a quarter.

Solution: We can use 1 coin, or 2 coins, or 3 coins, or 4 coins. Since all sums are different, the answer is C(4, 1) + C(4, 2) + C(4, 3) + C(4, 4) = 4 + 6 + 4 + 1 =15

EX 6 – Determine the number of ways a delegation of 4 students can be selected from 12

students if Henry and Scott cannot serve on the delegation together.

Solution: We use the BACKDOOR approach. 1) If any 4 students can serve, there would be C(12, 4) or 495 possibilities.

2) If Henry and Scott must be in the delegation, there would be C(10, 2) or 45 possibilities for the other 2 members.

Therefore, C(12, 4) – C(10, 2) = 450

26

EX 7 – A bag contains 6 black balls, 7 white balls, and 8 green balls. a) Determine the number of ways Zeke can create groups of 4 black or 4 white balls.

Solution: C(6, 4) + C(7, 4) = 15 + 35 = 50

b) Determine the number of ways Zeke can create groups of 4 black and 4 white balls.

Solution: C(6, 4)C(7, 4) = (15)(35) = 525

c) Determine the number of ways Zeke can create groups of 4 balls all the same color.

Solution: C(6, 4) + C(7, 4) + C(8, 4) = 15 + 35 + 70 = 120

d) Determine the number of ways Zeke can create groups of 4 balls of each color.

Solution: C(6, 4)C(7, 4)C(8, 4) = (15)(35)(70) = 36,750

EX 8 – A ternary word is a number that consists only of 0’s, 1’s and 2’s, where 0 could be the

first digit. Determine the number of ternary words of length 12 that have exactly seven 0’s, three 1’s, and two 2’s.

Solution: There are C(12, 7) ways to choose 7 of the 12 positions for 0’s, leaving C(5, 3)

ways to choose 3 of the remaining 5 positions for the 1’s, and C(2, 2) ways to choose 2 of the remaining 2 positions for the 2’s.

C(12, 7)C(5, 3)C(2, 2) = (792)(10)(1) = 7920 Alternate Solution: One can also view this as permuting the ternary digits 000000011122.

There are 12! 79207!3!2!

ways to do so.

EX 9 – Determine the number of ways 12 people can be partitioned into 3 groups of 4 people

each. Solution: Initially, this problem seems straightforward: C(12, 4)C(8, 4)C(4, 4) = 34,650;

however, the size in each group is the same. Consider the three groups to be A, B, and C. The groups could be selected as ABC, ACB, BAC, BCA, CAB, or CBA, yet looking at the three groups, they are all the same. Therefore, we must divide

by the number of ways groups could be ordered: C 12, 4 C 8, 4 C 4, 45775

3!

27

1.5 – Combinations – Problem Set 1. Expand and simplify each of the following. a) C(7, 2) b) P(4, 2)C(7, 4) c) C(12, 12) 2. Expand and simplify each of the following. Express your answers with no factorials. a) C(n, 1) b) C(n, n – 2) c) C(n + 2, n – 1) 3. From 250 boys and 100 girls, determine the number of committees that can be chosen

consisting of 2 boys and 1 girl. 4. For his starting defensive line this year, a football coach can choose from 3 candidates for

nose guard and 11 candidates for the other 4 linemen. Determine the number of different combinations which are possible to select.

5. From a group of 7 people, determine the number of committees that can be formed if there

must be at least 3 people on the committee. 6. Danielle must answer exactly 7 out of 10 questions on an English exam, including at least

3 out of the first 4 questions. Determine how many different combinations of questions she can choose.

7. An ace, a king, a queen, a jack, a 10, and a 9 are to be divided into two groups of three

each. Determine the number of ways the groups can be formed. 8. For this season’s starting basketball lineup, the coach can choose either of two players for

center, from 5 possibilities for 2 forwards, and from 6 possibilities for 2 guards. However, forwards Jenna and Kim cannot be in the lineup at the same time. Determine the number of possible starting lineups.

9. A bridge hand consists of 13 cards from a standard 52-card deck. Determine the number of

ways a person can get a bridge hand consisting of 4 aces, 3 kings, 2 queens, and 4 other cards that are either 3’s, 4’s, or 5’s.

28

10. The Key Club consists of 5 men and 6 women. a) Determine the number of ways a committee of 3 people can be formed.

b) Determine the number of ways a committee consisting of 3 men and 4 women can be formed.

c) Determine the number of ways a committee of 6 people can be formed if a certain pair of women refuse to serve on the same committee.

d) Determine the number ways a committee of 4 men and 3 women can be formed if 2 of the men refuse to serve on the same committee.

11. Determine the number of ways a team of 3 can be chosen from 8 people if: a) one particular person is always included. b) one particular person is always excluded. 12. Craig places one penny, one nickel, one dime, and one quarter on his desk. Determine how

many positive amounts of money Kathie can create from the coins on Craig’s desk. 13. Determine the number of ways a person get a bridge hand of 13 cards if the hand must

consist of at least 3 aces and at least 3 kings. Express your answer in terms of combinations.

14. If 6 coins are tossed together, determine the number of ways that: a) all fall heads. b) exactly two fall heads. c) at least four fall heads. 15. Determine the number of ways two numbers whose sum is even can be chosen, without

repetition, from the numbers 1, 2, 3, 8, 9, 10. 16. Determine the number of ways a debating team of three students can be chosen from 9

candidates. 17. Determine the number of 13-card hands consisting of 5 hearts and 8 spades that can be

formed from a standard 52-card deck.

29

18. In a volleyball mixed league, three men and three women must be on the court at one time. If there are 13 men and 11 women on the team, determine the number of combinations of players that can be on the court at any one time.

19. From a group of 20 sophomores and 10 freshmen, determine the number of committees of

4 sophomores and 3 freshmen are possible if at least one of 3 specific sophomores must be on the committee.

20. A hostess selects 6 luncheon guests from 10 women. If she must avoid having a particular

two of the women both selected. Determine the number of ways the guests can be chosen. 21. From a suit of 13 cards, determine the number of ways a group of 5 cards can be selected

that includes the King, or the Jack, or both. 22. From a group of 7 different books, determine the number of ways a) exactly 4 books can be chosen. b) at least 4 books can be chosen. c) at most 4 books can be chosen. 23. Determine the number of 13-card hands that can be dealt from a standard 52-card deck

which: a) have exactly 10 spades. b) have exactly 11 cards from any particular suit. 24. Given a bag containing 4 green balls and 5 red balls, determine the number of ways 3 balls

can be selected such that: a) 2 are green and 1 is red. b) all 3 must be the same color. 25. Determine the number of ways at least 5 students can be selected from 7 students.

30

1.6 – Binomial and Multinomial Theorems 1.6(A) Summation Symbol

Suppose we had this expression: 1! + 2! + 3! + 4! + 5! + 6! + 7! The writing could be shortened to 1! + 2! + 3! + … + 7! For reasons of space and convenience, mathematicians constantly invent shorter ways of

expressing ideas. We now present the SUMMATION SYMBOL as such a device. The symbol “ ” mathematically represents “the sum of…”. is the Greek letter known as sigma. Some textbooks refer to the ‘summation symbol’ as ‘sigma notation.’

The expression 1! + 2! + 3! + 4! + 5! + 6! + 7! can be expressed as:

7

n 1n!

In general, we substitute into the expression consecutive integers beginning with the number

below and ending with the number above. We then find the sum of all terms.

EX 1 – Determine 8

k 32k 1

Solution: Substitute consecutive integers beginning with 3 and ending with 8. If k = 3, 2k 1 2 3 1 7 If k = 4, 2k 1 2 4 1 9 If k = 5, 2k 1 2 5 1 11 If k = 6, 2k 1 2 6 1 13 If k = 7, 2k 1 2 7 1 15 If k = 8, 2k 1 2 7 1 17

Therefore, 8

k 32k 1 7 9 11 13 15 17 72

EX 2 – 5

n 1n n 2 1 3 2 4 3 5 4 6 5 7 85

31

1.6(B) The Binomial Theorem EX 3 – Expand and simplify 5x y

Solution: 5 0 4 1 3 2 2 3 1 4 0 55 5 5 5 5 5x y x y x y x y x y x y

0 1 2 3 4 5

5 4 3 2 2 3 4 5x 5x y 10x y 10x y 5xy y EX 4 – Expand and simplify 3x 2y

Solution: 0 1 2 33 2 1 03 3 3 3x 2y x 2y x 2y x 2y

0 1 2 3

3 2 2 3x 6x y 12xy 8y EX 5 – Determine the term involving 3x in the expansion of 5x 4 .

Solution: 3 2 35! x 4 160x3!2!

3 2 2 3x 6x y 12xy 8y

The Binomial Theorem

For any positive integer n, n

n n r r

r 0

na b a b

r

Another way to express C(n, r) is nr

. In this section, we will be

using the symbol nr

to represent the combination of “n” items

taken “r” at a time.

32

1.6(C) The Multinomial Theorem

How would you expand 3a b c ? This could be done by considering (a + b) as one term and applying the Binomial Theorem: 33 3 2 2 3a b c a b c a b 3 a b c 3 a b c c Now apply the Binomial Theorem again to 3a b and 2a b : 3 3 2 2 3 2 2 2 2 3a b c a 3a b 3ab b 3a c 6abc 3b c 3ac 3bc c Rearranging the terms, we obtain: 3 3 3 3 2 2 2 2 2 2a b c a b c 3a b 3a c 3b c 3ab 3ac 3bc 6abc

We could use the same process to expand 4 20a b c d or a b c d e . However, if we apply our knowledge of permutations, we can find a shortcut. If we expand 4a b c d , we would obtain the term 2a bc in each of the following ways:

aabc, aacb, abac, abca, acab, acba, baac, baca, bcaa, caab, caba, cbaa

Therefore, one of the terms in the expansion of 4a b c d must be 212a bc.

Recall from Section 1.4:

Permutation Theorem 2 - Repetition

The number of permutations of “n” objects of which “p” are alike, “q” are alike, “r” are alike, etc., is:

n!p!q!r!...

33

We can apply this to form the Multinomial Theorem: EX 6 – Apply the Multinomial Theorem to expand 4a b c .

Solution: 4 4 4 3 0 3 04! 4! 4! 4! 4!a b c a bc a b c4!0!0! 0!4!0! 0!0!4! 3!1!0! 3!0!1!

0 3 2 2 0 2 0 2 0 2 2 24! 4! 4! 4! 4!a b c a b c a b c a b c a bc0!3!1! 2!2!0! 2!0!2! 0!2!2! 2!1!1!

2 2 3 0 0 3 0 34! 4! 4! 4! 4!ab c abc ab c ab c a bc1!2!1! 1!1!2! 1!3!0! 1!0!3! 0!1!3!

Simplifying, we obtain:

4 4 4 4 3 3 3 2 2 2 2 2 2

2 2 2 3 3 3

a b c a b c 4a b 4a c 4b c 6a b 6a c 6b c

12a bc 12ab c 12abc 4ab 4ac 4bc

Note: While not necessary, terms with an exponent of 0 (and 0!) are included in this example for the purpose of understanding.

EX 7 – Determine the numerical value of 3

3

x 0x 1

Solution: 3 3 3 30 1 1 1 2 1 3 1 1 8 27 64 100

The Multinomial Theorem For any positive integer n, each term of the expansion na b c ... is of the form:

p q rn! a b c ...p!q!r!...

where p + q + r +… = n.

34

EX 8 – Expand and simplify 5

2 3x .y

Solution: 0 1 2 3

5 4 3 22 2 2 25 5 5 53 3 3 3x x x x0 1 2 3y y y y

4 5

1 02 25 53 3x x4 5y y

8 6 4 2

102 3 4 5

15x 90x 270x 405x 243xy y y y y

EX 9 – Determine which term in the expansion of 92 1x 7x is 9,882,516 for any nonzero

value of x. Solution: Since we are looking for a constant term, the exponent on x for this term in the

expansion must be 0. Since term has the form 9 a a2 19x 7x .

a

Looking

only at the x term and using properties of exponents, 18 2a a 0x x x

18 3a 0x x 18 3a 0 a 6 , which is the seventh term. Note that

6

329 7x 9,882,5166 x

.

EX 10 – Expand and simplify 3a b c d

Solution: 33 3 3 2 2 2 2 23! 3! 3! 3! 3!a b c d a b a c a d b c b d2!1! 2!1! 2!1! 2!1! 2!1!

2 2 22 2 23! 3! 3! 3! 3! 3!c d ab a c ad b c bd2!1! 2!1! 2!1! 1!2! 1!2! 1!2!

2

3 3 3 3 2 2 2 2 2 2 2 2

2 2 2 2

3! 3! 3! 3! 3!c d ab c abd a c d b c d1!2! 1!1!1! 1!1!1! 1!1!1! 1!1!1!

a b c d 3a b 3a c 3a d 3b c 3b d 3c d 3ab 3ac3ad 3bc 3bd 3cd 6abc 6abd 6acd 6bcd

35

EX 11 – Determine the number of terms in the expansion of 5a b c d e .

Solution: There are 5 terms with coefficient 5!5!0!0!0!0!

.

There are 20 terms with coefficient 5!4!1!

.

There are 20 terms with coefficient 5!3!2!

.

There are 30 terms with coefficient 5! .3!1!1!

There are 30 terms with coefficient 5! .2!2!1!

There are 20 terms with coefficient 5! .2!1!1!1!

There is 1 term with coefficient is 5!1!1!1!1!1!

.

Therefore, there are 126 terms

Alternative Solution: 5 5 1 9

1265 1 4

Consider this problem: Determine the sum of the coefficients in the expansion of 2a b .

Solution: 2 2 2a b a 2ab b . The sum of the coefficients is 1 + 2 + 1 = 4. Notice that substituting a = b = 1 in the original expression results in 21 1 4

The number of terms in the expansion of n1 2 3 ra a a ... a is:

n r 1r 1

The sum of the coefficients in the expansion of a binomial/multinomial expression can be found by substituting a 1 for each variable and evaluating the expression.

36

EX 12 – Determine the sum of the coefficients in the expansion of 7a b c d .

Solution: 71 1 1 1 16,384

EX 13 – Given the expansion of 15x 2y , a) Determine the tenth term in the expansion.

Solution: Recall from the Binomial Theorem in the first term, r = 0. Therefore, for the tenth

term, r = 9: 6 9 6 915x 2y 2,562,560x y

9

b) Determine the sum of the coefficients in the expansion.

Solution: 151 2 1 1

EX 14 – Consider the expansion of 52a 3b 2c . a) Determine the sum of the coefficients in the expansion of 52a 3b 2c .

Solution: 52 1 3 1 2 1 243

b) Determine the coefficient of 2 3a c in the expansion of 52a 3b 2c .

Solution: 2 0 3 2 35! 2a 3b 2c 320a bc 3202!0!3!

c) Determine the term with 2 2a bc in the expansion of 52a 3b 2c .

Solution: 2 1 2 2 25! 2a 3b 2c 1440a bc2!1!2!

37

1.6 – Binomial and Multinomial Theorems – Problem Set

1. Expand and evaluate 3

k 0k k 3

.

2. Expand and evaluate 5

x 1

xx 1 .

3. Expand and evaluate 4

k 1

k k 12

4. Expand and simplify 4

2xy

.

5. Expand and simplify 3a b . 6. Determine the seventh term of 102x y .

7. Determine the fourth term of 51x

x

.

8. Determine the eleventh term of 123x 2y .

9. Determine the constant term of 12

2 1x .x

10. Determine the numerical coefficient of the twelfth term of 14k 1 .

38

11. Determine the coefficient of 3 7a b in the expansion of 10a b . 12. Determine the sum of the coefficients in the expansion of 112a b . 13. Determine the sum of the coefficients in the expansion of 52a 2b 4c . 14. Expand and simplify 32a x c . 15. Given 6x y z w : a) Determine the coefficient of the term involving 2 3x yz . b) Determine the sum of the coefficients of the terms involving 3 3 3x y and xyzw . c) Determine the sum of the coefficients in the expansion of the expression. d) Determine the number of terms in the expansion of the expression. 16. Determine the coefficient of 4x in the expansion of 52x x 2 . 17. Given 52x 2y :

a) Determine the coefficient of the term involving 4 3x y . b) Determine the number of terms in the expansion of the expression. c) Determine the sum of the coefficients in the expansion of the expression. d) Determine the fifth term in the expansion of the expression. 18. Given 4x 2y z : a) Determine the sum of the coefficients of the terms involving 2 2 2x yz and y z . b) Determine the number of terms in the expansion of the expression. c) Determine the sum of the coefficients in the expansion of the expression.

39

1.7 – Combinatorial Equations 1.7(A) Expressions and Equations with Factorials

Many expressions with factorials can be simplified to an algebraic expression with no factorials.

Consider the expression

n! :n 1 !

n! n n 1 n 2 n 3 n 4 ... 1

n 1 ! n 1 n 2 n 3 n 4 ... 1

Therefore,

n n 1 n 2 n 3 n 4 ... 1n! nn 1 ! n 1 n 2 n 3 n 4 ... 1

For an alternate (and simpler) solution, explicitly list the beginning factors of the ‘larger’ factorial expression until you are left with a factorial that matches the smaller expression.

Reconsider the expression

n! .n 1 !

We know that n!has one more factor than n 1 ! since n is one larger than n 1 . Thus, we can rewrite n! as n n 1 !

n n 1 !n! nn 1 ! n 1 !

This is how solutions will be given in this section.

EX 1 – Expand and simplify

n 1 !.

n 2 !

Solution:

n 1 n n 1 n 2 !n 1 n n 1

n 2 !

Note: In this section, we will consider the domain of all variables to be whole numbers.

40

EX 2 – Expand and simplify

3n !.

3n 2 !

Solution: 3n 3n 1 3n 2 !

3n n 13n 2 !

EX 3 – Expand and simplify P(n, 2).

Solution: n n 1 n 2 !n!P n, 2 n n 1

n 2 ! n 2 !

EX 4 – Expand and simplify n 12

.

Solution:

n 1 n 1 ! n 1 ! n 1 n 2 n 3 ! n 1 n 22 n 3 !2! n 3 !2! 2n 1 2 !2!

EX 5 – Determine n if

2n 6 !n 16n 28

n 4 !

Solution:

2n 6 n 5 n 4 !

n 16n 28n 4 !

2n 6 n 5 n 16n 28

2 2

2

n 11n 30 n 16n 282n 5n 2 0

2n 1 n 2 01 1n , 2 reject n 22 2

41

EX 6 – Determine n if n 2 !

19n 1 !

Solution:

n 2 n 1 !

19n 1 !

n 2 19 n = 17

EX 7 – Determine n if 5n 9 ! 15n 7 ! 6

Solution:

5n 9 ! 1

5n 7 5n 8 5n 9 ! 6

2

2

1 15n 7 5n 8 6

25n 75n 56 625n 75n 50 025 n 1 n 2 0

n 1, 2 reject 1 n 2

EX 8 – Determine n if

2n 2 !8n 52n

n 1 !

Solution:

2n 2 n 1 n n 1 !

8n 52nn 1 !

2

2

2

n 2 n 1 n 8n 52n

n 2 n 1 n n 8n 52

Since n 1: n 2 n 1 8n 52

n 3n 2 8n 52n 5n 50 0

n 10 n 5 0n 5,10 reject 5 n 10

42

1.7(B) Combinatorial Equations

EX 9 – Determine n3

if P n, 7 30P n, 5 .

Solution:

n! 30n!n 7 ! n 5 !

2

2

n n 1 n 2 n 3 n 4 n 5 n 6 30n n 1 n 2 n 3 n 4

n 5 n 6 30

n 11n 30 30n 11n 0

n n 11 0n 0,11 reject 0 n 11

11 11! 1653 11 3 !3!

EX 10 – Determine P(n, 2) if n 1 n

3 73 2

.

Solution:

3 n 1 ! 7n!

n 2 !2!n 1 3 !3!

n 1 ! 7n!n 2 ! n 2 !

n 1 n n 1 7n n 1n 1 7

n 66!P 6, 2 30

6 2 !

43

EX 11 – Determine P(n, 1) if n

P n 1, 2 922

Solution:

n 1 !n! 92

n 2 !2! n 3 !

2 2

2

n n 1n 1 n 2 92

2n n 1 2 n 1 n 2 184

n n 2n 6n 4 1843n 7n 180 0

n 9 3n 20 020 20n 9, reject n 93 3

9!P 9,1 99 1 !

EX 12 – Determine n3

if P n 1, 2 P n 3, 2 30 .

Solution:

n 1 ! n 3 !

30n 1 2 ! n 3 2 !

2 2

n 1 n 2 n 3 n 4 30

n 1 ! n 3 !30

n 3 ! n 5 !

n 3n 2 n 7n 12 304n 40

n 1010 10! 1203 10 3 !3!

44

EX 13 – Determine P 6, r if

7r 59 18r

.

Solution:

7!7 r !r! 5

9! 189 r !r!

2

2

7! 9 r !r! 59! 7 r !r! 18

9 r 8 r 572 18

9 r 8 r 20

72 17r r 20r 17r 52 0

r 13 r 4 0r 4,13 reject 13 r 4

6!P 6, 4 3606 4 !

EX 14 – Determine 21n

if P(n, 3) = 17P(n, 2).

Solution:

n! 17n!n 3 ! n 2 !

n n 1 n 2 17n n 1n 2 17

n 1921 21! 21019 21 19 !2!

45

1.7 – Combinatorial Equations – Problem Set

1. Expand and simplify

n!n 2 !

.

2. Expand and simplify P n, 3 .

3. Expand and simplify n

.4

4. Expand and simplify

nk 1

.n 1k

5. Determine n if

2n 3 !n 10n 4

n 1 !

.

6. Determine n if

22n 2 !3n 9n 30

2n !

.

7. Determine n if

2

2

n 4 !18

n 2 n 5 !

.

46

8. Determine n if

2n! 14n 1 ! n 1 ! 15

.

9. Determine n if 3n 1 !

123n 1 !

.

10. Determine n if

2n 1 ! 3n ! n 1 !50

2n ! 3n 1 ! 2n!

.

11. Determine n2

if P(n, 5) = 14P(n, 4).

12. Determine 19n

if 3P n, 5 68P n 2, 4 .

13. Determine P(n, 2) if n n 2

4 972 2

.

14. Determine n3

if 5P(n, 3) = 2P(n – 1, 4).

15. Determine 12n

if P(n, 4) = 3024.

47

16. Determine P(6, n) if n n 1

42 2

.

17. Determine 7n

if P(n + 1, 3) = P(n, 4).

18. Determine P(n, 2) if n n n

5n1 2 3

.

19. Determine 152n

if 10P(n, 1) + 2P(n, 2) = P(n, 3).

20. Determine P(n, 4) if

n 2n 1 !4

2n 2 !

.

48

1.8 – Combinatorics – Problem Set This is by far the most challenging section of Chapter 1. You computed permutations in the ‘permutations’ sections, combinations in the ‘combinations’ section, overlapping possibilities in the ‘overlapping possibilities’ section, etc. You now have many of the tools necessary to complete combinatorial problems. Now comes the challenge: selecting the appropriate method(s) given a random combinatorial problem. We begin that challenge with the first problem below. 1. State whether each of the following represents a permutation or a combination. a) a hand of 13 cards drawn from a deck of cards b) five pictures placed in a row c) three paperbacks picked from a rack d) a team of 9 players chosen from a group of 20 e) the arrangement of the letters in the word OCTOBER f) three colors selected from red, orange, yellow, green, blue, and violet g) announcing the starting lineup of 5 basketball players h) 1st thru 3rd place in a race of 8 runners 2. a) Five cards consisting of an Ace, King, Queen, Jack, and 10 are divided into two groups,

one containing 2 cards and one containing 3 cards. Determine the number of ways groups can be created.

b) Six cards consisting of an Ace, King, Queen, Jack, 10, and 9 are divided into two groups, each containing 3 cards. Determine the number of ways groups can be created.

3. Determine the twelfth term in the expansion of 14k 1 . 4. The captain of a baseball team assigns himself to the 4th place in the batting order.

Determine the number of ways he can assign the remaining places to his 8 teammates if just 3 men are considered eligible for the 1st position.

5. Determine how many ways a hand consisting of exactly 5 spades, exactly 6 hearts, and 2

other cards can be selected from a standard 52-card deck. 6. Given eight students: a) Determine the number of ways all eight can be lined up on the 50-yard line. b) Determine the number of ways all eight can be arranged in a circle.

49

7. Determine how many six-digit numbers can be created from the digits 3, 3, 4, 4, 4, 5, 6 if all repeated digits are used.

8. Expand and simplify 6! .3!2!

9. Determine 11n

if 2P(n + 1, 2) = 108 + P(n, 2).

10. A license plate consists of two different consonants followed by 6 different digits with the

first digit non-zero. Determine the number of possible license plates. 11. In an examination with 13 questions, a student is to answer 10, two or three of which must

be chosen from the first three questions. Determine the number of ways the student can choose the 10 questions.

12. A survey of 135 students asked whether they like basketball or football. 64 liked

basketball, 75 liked football, and 8 liked neither basketball nor football. Determine the number of students who liked both basketball and football.

13. Determine the term involving 4 4x y in the expansion of 822x y 3 . 14. Determine the number of possible license plates if the first two characters must be letters,

at least one of which is a vowel, and the last four characters are digits, the first of which is non-zero.

15. Seven boys and five girls are members of a summer tennis team.

a) Determine the number of ways to form a doubles team (2 players) for the next tournament.

b) Determine the number of ways to form a mixed doubles team (one boy and one girl) for the next tournament.

c) Determine the number of ways a mixed doubles match can be arranged among team members.

50

16. There are 5 roads connecting A and B and 2 roads connecting B and C. Determine the number of ways one can do a round trip from A to C and back to A.

17. Determine n if n 2 ! 42n! 18. Determine how many numbers less than 500 can be created using the digits 1, 2, 3, 4, 5, 6,

and 7. 19. Determine the number of products that can be formed from the numbers 2, 3, 4, 5, 6 taking

two or more at a time. 20. Determine the number of ways 12 different colored beads can be arranged on a bracelet.

21. Determine n2

if n 2

P n, 2 242

.

22. A host is seating himself and 4 guests. a) Determine the number of ways all can be seated around a circular table. b) Determine the number of ways the guests can be seated after the host seats himself.

23. Expand and simplify 6!4!

.

24. Five coins are tossed. Determine the number of ways they land so at least two heads

appear. 25. Determine the number of ways 10 people can be divided into 2 groups of 3 people and 2

groups of 2 people.

51

26. The cafeteria sold pizza, hamburgers, and tacos for today’s lunch. A total of 30 students were in the cafeteria for period 5B lunch. 14 chose pizza, 11 chose a hamburger, 11 chose a taco, 8 chose pizza and a hamburger, 6 chose a hamburger and a taco, 7 chose pizza and a taco, and 5 chose all three. Determine the number of students who brought their lunch from home (purchased none of the three).

27. From the digits 1, 2, 3, 4, 5, 6, 7, 8, and 9, determine how many numbers of 7 different

digits each can be formed in which odd and even digits alternate. 28. From the letters in the word DECIMAL, arrangements of 5 different letters are formed. a) Determine the number of arrangements that can be formed. b) Determine the number of arrangements that begin and end with a consonant. c) Determine the number of arrangements that have A or C in the middle. d) Determine the number of arrangements that have consonants and vowels alternating. 29. Determine the sum of the coefficients in the expansion of 7a b c d . 30. a) Determine the number of different hands of 13 cards each can be made from a standard

52-card deck. Leave your answer in combination form. b) Determine the number of ways 4 groups of 13 cards each can be created.

31. A gym lock has a combination which includes three numbers from zero to 36 inclusive.

Determine the number of combinations that include at least one 24 in the combination. 32. From 3 sophomores, 4 juniors, and 5 seniors, determine the number of different 5-person

committees that can be formed consisting of 1 sophomore, 1 junior, and 3 seniors. 33. Determine how many numbers of at least two different digits can be formed by using the

digits from the set {1, 2, 3, 4, 5}.

34. Determine n if

n 1 !20n 20

n 2 !

52

35. A group of students were surveyed as to whether they like drama club or music club. 24 students like drama, 22 students like music, 15 like both music and drama, and 12 like neither music nor drama. Determine the number of students that answered the survey.

36. Determine the number of ways seven students can be divided into 3 groups of 2. 37. Expand and simplify 5x 3y . 38. Determine the number of different signals, each consisting of 6 flags hung in a vertical

line, that can be formed from 3 blue flags, 2 green flags, and 1 gray flag.

39. Determine n5

if n 1

P n, 4 303

, determine n.

40. Determine the coefficient of 2 2a b c in the expansion of 6a b c 2 . 41. Determine the number of different necklaces that can be assembled if nine beads of

different colors are used. 42. 3 freshmen, 5 sophomores, 4 juniors and 2 seniors have been nominated to serve on a

student advisory committee. Determine the number of possible committees if: a) the committee is to consist of any four persons. b) the committee is to consist of one freshman, one sophomore, one junior, and one senior.

c) the committee is to consist of two persons: one freshman or sophomore and one junior or senior.

d) the committee is to consist of three persons from different classes. 43. Determine the number of ways a committee of 5 can be chosen from a group of 10.

53

44. The foreign language department offers French, Spanish, and German classes. 100 foreign language students answered a survey to see which class(es) they had taken. 52 replied French, 63 replied Spanish, 25 replied German, 4 replied all three, 24 replied French and Spanish, and 11 replied Spanish and German. Determine the number of students who replied French and German.

45. Determine the number of ways 6 students can be seated in a row if 3 of the 6 insist on

sitting next to each other. 46. Determine the number of ways to select five people from a group of seven people and seat

them in a row of five chairs. 47. There are 3 freshmen, 2 sophomores, 2 juniors, and 3 seniors eligible for a 5-person

calculator team competition. Determine the number of possible teams that include at least one freshman.

48. Five boys and five girls were nominated for the homecoming court. Determine the number

of ways a king, a queen, and a court of 2 students can be selected from those nominated. 49. Using the cards from a standard 52-card deck, determine the number of ways a player can

receive a 5-card hand which has at least two Aces. 50. Determine how many permutations exist using all letters of the word MATHEMATICS. 51. Among 18 students on a college dorm floor, 7 are majoring in mathematics, 10 are

majoring in business, and 10 are majoring in computer science. 3 students are double majoring in mathematics and business, 4 in mathematics and computer science, and 5 in business and computer science. 2 students are not majoring in any of the three. Determine how many of the 18 students are triple majoring in mathematics, business and computer science.

54

52. The digits 3 through 8 are used to make four-digit numbers. a) Determine how many numbers can be made if repetition is allowed. b) Determine how many numbers can be made if repetition is not allowed.

c) Determine how many numbers can be made using the digit 5 at least once, if repetition is allowed.

d) Determine how many numbers can be made using the digit 5 at least once, if repetition is not allowed.

53. Expand and simplify n 3 !n 1 !

54. There are 8 boys and 12 girls in the drama club. Determine the number of ways a cast of

five can be selected if: a) there must be at least 2 boys. b) there must be at least 2 girls. c) there must be more girls than boys. 55. Determine the number of ways 7 people can be seated in a circle if 3 people insist upon

sitting next to each other. 56. Determine the fourth term in the expansion of 52y 3z .

57. Expand and simplify 6

n 3n n 2

58. In a club raffle, tickets are numbered 1 to 25. 6 winning tickets are drawn. Determine the

number of ways this can be done if: a) the selection contains the numbers 1 and 14. b) the selection contains the same number of odd numbers as even numbers. 59. The 44-member baseball team attended the school’s volleyball match to support them. The

coach of the baseball team pre-ordered hot dogs from the concession stand for his team. 5 baseball team members did not want a hot dog. 28 baseball team members ordered a hot dog with ketchup, 20 with mustard, 10 with ketchup and mustard, 8 with mustard and relish, 11 with ketchup and relish, and 6 with ketchup, mustard, and relish. Determine the number of members of the baseball team who wanted their hot dog with only relish.

55

60. If a bridge hand consists of 13 cards from a standard 52-card deck, determine the number of ways a player can get a hand consisting only of Aces, Kings, Queens, Jacks, or 10s.

61. A mixed basketball team is made of 3 males and 2 females. If there are 13 males and 10

females eligible for the team, determine the number of possible teams. 62. Determine the number of ways can a person distribute a ball, a bat, and a mitt: a) to 5 boys. b) to 5 boys if no boy is to receive more than one. 63. A bag contains 5 red, 6 blue, and 4 yellow marbles. Three are drawn at random without

replacement. Determine the number of ways a) all are blue. b) all are the same color. c) all are different colors. 64. Given the digits 1, 1, 2, 2, and 2: a) Determine how many different five-digit numbers can be created.

b) Determine how many of the numbers formed from the given digits are greater than 12,000 and less than 13,000.

65. Determine the number of ways the letters of the word WOLVES can be arranged if the first

letter must be a vowel. 66. During the basketball season, a conference of 10 teams plays each team twice, once home

and once away. Determine the total number of games in the conference season. 67. A group of 71 coffee drinkers were asked what they added to their coffee. 36 responded

cream and 55 responded sugar. If all who were surveyed responded cream and/or sugar, determine the number of coffee drinkers who add both cream and sugar to their coffee.

68. The Super Sundae Ice Cream Store sells a mega-sundae, consisting of any 2 flavors of ice

cream, 2 kinds of syrup, and 3 toppings. If the store had 10 flavors of ice cream, 4 kinds of syrup, and 6 varieties of toppings, determine the number of different mega-sundaes the store can sell.

56

69. From 7 red balls, 3 white balls, and 4 blue balls, Patty draws a ball at random three times, each time replacing the ball drawn. Determine the number of ways Patty can draw a red ball, white ball, and a blue ball in that order.

70. Determine n 1

2

if P(2n, 3) = 84n.

71. Expand and simplify 5

n 32n 1

.

72. When it comes to video games, 5 people prefer League of Legends, 5 prefer Counter-

Strike, and 5 prefer Minecraft. Determine the number of ways 4 of these people can be selected such that each preferred game is represented.

73. In the expansion of 42a b c d , determine the term containing 4abc .

74. Expand and simplify 5! .2!0!

75. Determine P(n, 3) if n

153.2

76. Determine the term which contains 12

7yx

when 83

22x yy 2x

is expanded.

77. In rolling a die twelve times, Matt rolled three 6’s, three 5’s, two 4’s, two 3’s, one 2 and

one 1. Determine the number of distinct orders this could have happened.

57

78. David leaves home for work, which is 8 blocks west and 9 blocks north of his home, and travels 17 blocks along a lattice grid in full 1-block increments. There is a fast food restaurant, which he loves, that is 4 blocks west and 3 blocks north of his home. Determine the number of ways David can go to work, stopping at the fast food restaurant for breakfast.

79. There are 10 students in the Gaming Club. a) Determine the number of ways a President and Vice-President can be selected.

b) From the remaining 8 students, a committee of 3 is to be formed to research the latest games. Determine the number of ways this committee could be formed.

80. The nine members of the city council convene every other Tuesday. There are four

Democrats, three Republicans, and two independents on the council. Before each meeting begins, they shake hands; however, Republicans and Democrats do not shake hands with each other. Determine the number of handshakes that occur before each meeting.

81. If a bridge hand consists of 13 cards from a standard 52-card deck, determine the number

of ways a player can get a hand including at least 3 Aces and at least 3 kings. Express your answer in terms of combinations.

82. In a school of 320 students, 85 are in the band, 200 are on a sports team, and 60 are in both

sports and the band. Determine the number of students who are in neither band nor on a sports team.

83. Determine the number of ways 7 officers can be divided into two groups of 2 and a group

of 3 if a specific officer named Ken must be in a group of 2. 84. You are laying out silverware, in a row, to dry. Determine the number of possible

arrangements with 11 pieces of silverware from the same pattern consisting of 3 teaspoons, 3 forks, 3 knives, and 2 soup spoons.

85. Determine the number of ways 11 Key Club members can be arranged in a line for the

yearbook photo if the president must be in the middle, the vice-president must be next to her, and the other three officers must be in the group with the president and vice-president.

58

86. From a standard 52-card deck, determine the number of possible 5-card hands if: a) there are only aces or face cards. b) there are only numbered cards (no letters). c) there are two clubs and three diamonds. d) there are at least 4 red cards. e) there are exactly three 5’s. f) there are exactly 2 queens and exactly 2 hearts. 87. Expand and simplify P(8, 5). 88. Determine the number of ways 4 men and 4 women can be seated around a table with men

and women alternating. 89. Determine how many different numbers of at most three digits can be formed from the

digits 4, 5, 6, if: a) no repetition is allowed b) repetition is allowed.

90. Determine n if 2n! 5n 11n

n 3 !

91. An online gaming store has 8 sports games and 12 adventure games on sale. Determine the

number of ways 5 games can be purchased if you must purchase at least 2 sports games. 92. Determine n if P(n, 2) = 2C(n, 4)

93. Determine the constant term in the expansion of 4

22

23x .x

94. The local steakhouse offers a choice of three side dishes with each steak. They offer 6

different vegetables and 4 different potato sides. Nancy wants at least one vegetable and at least one potato side. Determine the number of ways Nancy can order her three side dishes.

59

95. Given 7 flags of different colors, determine the number of different signals that can be generated if a signal requires at least four flags, each below the other.

96. Determine the number of different passenger tickets Amtrak will need for use on a railroad

line that has twenty stations. 97. Determine the number of permutations that can be made from the letters of word

“assassination” taken all together. 98. Expand and simplify

92

n 5n 6n

.

99. Determine how many numbers between 500 and 1000 which have exactly one 8 in the

number. 100. 150 college freshmen were interviewed. 85 were registered for a math class, 70 were

registered for an English class, and 50 were registered for both a math class and English class.

a) Determine the number of college freshmen who registered for only a math class. b) Determine the number of college freshmen who registered for neither a math class nor

English class.

60

1.1 – Two Fundamental Principles – Problem Set – SOLUTIONS 1. 12 11 10 9 8 95, 040 2. a) 2 3 2 12 b) 2 4 4 32 3. 13 17 21 4, 641 4. 3 2 3 2 5 36 5. a) 26 26 10 10 10 10 6, 760, 000 b) 26 25 10 9 8 7 3, 276, 000 c) 26 25 9 9 8 7 2, 948, 400 6. 1 7 6 5 4 3 2 1 5, 040 7. 3 06 8. 4 5 6 77 7 7 7 960,400 9. 7 6 5 210 10. 26 10 10 10 26 10 10 10 26 7800 26 26 10 10 26 26 26 10 26 20, 280 26 26 26 17,576 7800 20, 280 17,576 45, 656 11. 1 5 1 10 5 55 12. 26 25 9 9 8 7 2,948, 400 13. 84 65,536

14. 36 36 36 15 11,664

60

15. a) 5 6 3 4 42 b) ABC: 5 6 , BAC: 3 4 , BCA: 6 4 , ACB: 4 2 , CAB: 4 5 , CBA: 2 3 30 12 24 8 20 6 100 c) ABA: 3 5 , ACA: 4 4 , BAB: 3 5 , BCB: 6 2 , CAC: 4 4 , CBC: 2 6 100 15 16 15 12 16 12 186 16. 3 4 5 3 4 3 5 4 5 107 17. a) 6 5 4 3 2 1 1 1 4 2880 b) The answer would be multiplied by 3 since there are 3 possible periods for lunch. 18. 3 5 6 3 5 4 3 6 4 5 6 4 342 19. a) 16 15 240 b) 16 16 256 20. a) 4 3 2 24 b) 4 4 4 64 c) 4 1 1 1 4

61

1.2 – Overlapping Possibilities – Problem Set – SOLUTIONS 1. n I C 107 x, n I 60, n C 43, n I C 9 n I C n I n C n I C 107 x 60 43 9 x 13 2. n M W x, n M 21, n W 17, n M W 12 n M W n M n W n M W x 21 17 12 26 3. n V C CC 18, n V 11, n C 10, n CC 8, n V C 3, n V CC 5, n CC C 5, n V C CC x a) 18 11 10 8 3 5 5 x x 2

b) 5 5 2 8 liked cookies and cream and at least one other flavor. Since 8 liked cookies and cream, there are zero participants who liked only cookies and cream.

4. n(divisible by 3) = 66, n(divisible by 5) = 40, n(divisible by 15) = 13 66 40 13 93 5. n S C 100 56 44, n S 18, n S C 10 Since 44 attended a least one class and 10 attended both classes, 44 – 10 = 34 attended

exactly one class. 6. 2x 5 x 7 21 68 150 x 21 2x 5 x 7 3x 2 3 21 2 61 7. n V R M 90, n V 60, n R 60, n M 70, n V R M 26, n V R 10 26 36, n V M 20 26 46, n R M 18 26 44 a) 60 36 46 26 4 b) 70 46 44 26 6 8. n P C B 135 x, n P 63, n C 61, n B 53, n C B 28, n P C 31, n P B 22, n P C B 15. a) 135 x 63 61 53 28 31 22 15 x 24 b) 63 31 22 15 25 9. n P G 24 4 20, n P 12, n G 13, n P G x 20 12 13 x x 5 10. [all numbers in thousands] n M 119, n I 131, n T 140, n M I T 16 n M I 25 16 41, n M T 19 16 35, n I T 27 16 43 a) 140 35 43 16 78 78,000 b) 41 43 35 2 16 87 87,000 11. 5 4 5 3 3 4 3 3 2 2 2 2 2 1 x 6 x 1

62

1.3 – Factorials and Permutations – Problem Set – SOLUTIONS 1. a) 4 3 2 1 24

b)

6 5 4 3 2 16! 1206 3 ! 3 2 1

c)

9 8 7 6 5 4 3 2 1

727 6 5 4 3 2 1

d) 8 7 6 5 4 3 2 1

286 5 4 3 2 1 2 1

e)

n n 1 n 2 n 3 n 4 ... 1n n 1 n 2

n 3 n 4 ... 1

f)

8 7 6 5 4 3 2 18! 568 2 ! 6 5 4 3 2 1

g)

n n 1 n 2 n 3 n 4 ... 1 n n 1 n 2 n 3 n 4 ... 1n n n 1

n 1 n 2 n 3 n 4 ... 1 n 2 n 3 n 4 ... 1

h)

4 3 2 14! 244 4 ! 1

i)

7 6 5 4 3 2 142

5 4 3 2 1 1

2. a)

12 11 10 9 8 7 6 5 4 3 2 112!P 12, 4 11,880

12 4 ! 8 7 6 5 4 3 2 1

b)

16 15 14 13 12 11 ... 3 2 116!P 16, 4 43,680

16 4 ! 12 11 ... 3 2 1

3.

11 10 9 8 7 6 ... 3 2 111!P 11, 5 55,440

11 5 ! 6 5 ... 3 2 1

4.

6 5 4 3 2 16!P 6, 3 120

6 3 ! 3 2 1

5. P 9, 2 P 9, 4 P 9, 6 P 9, 8 72 3024 60480 362880 426,456 6. P 5, 3 P 8, 4 60 1680 100,800 7. P 6, 3 P 6, 4 P 6, 5 120 360 720 1200

8.

10 9 8 7 6 5 4 3 2 110!P 10, 3 720

10 3 ! 7 6 5 4 3 2 1

63

1.4 – More on Permutations – Problem Set – SOLUTIONS

1. 15! 756,7565!5!5!

2. a) 9! 362,880 b) 5!4! 2880 c) 5!4! 2 5760 3. 3 32 6 2 1 8 3 1 5 4. a) 3!5! 720 b) 6 1 ! 2 240 5. 3 5! 360

6. 7! 25202!

7. 5 1 !12

2

8. P 10, 3 P 5, 3 660 9. a) 7! 5040 b) 3!4! 144 c) 3 5 4 3 2 1 2 720 d) 4 5 4 3 2 1 3 1440

10. 9! 12603!4!2!

11. 26 26 10 10 10 10 26 26 26 10 10 10 25 25 10 10 10 10 25 25 25 10 10 10 2,461,000 12. 8 1 ! 5040 13. 2!4! 48 14. 26 26 8 8 8 346,112

15. 13! 7159!4!

16. 2 5 4 3 120

17. 7 1 !360

2

18. 5 5 4 5 4 3 4 4 3 4 3 2 45 19. a) 3!5! 720 b) 3 5 4 3 2 1 4 1440 20. a) 5 1 ! 24 b) 4! 24

21. 5 1 !12

2

22. 5!4! 2880

23. 8! 4204!2!2!

24. 9! 1265!4!

64

1.5 – Combinations – Problem Set – SOLUTIONS

1. a)

7 67! 7! 21

7 2 !2! 5!2! 2 1

b)

7 6 5 44! 7! 4!7! 4204 2 ! 7 4 !4! 2!3!4! 2 1

c) 12! 12! 1

12 12 !12! 0!12!

2. a)

n n 1 !n! n

n 1 !1! n 1 !1

b)

n n 1 n 2 ! n n 1n!2! n 2 ! 2n n 2 ! n 2 !

c)

n 2 ! n 2 n 1 n n 1 ! n 2 n 1 n3! n 1 ! 6n 2 n 1 ! n 1 !

3. C 250, 2 C 100,1 3,112,500 4. C 3,1 C 11, 4 990 5. C 7, 3 C 7, 4 C 7, 5 C 7, 6 C 7, 7 99 6. C 4, 3 C 6, 4 C 4, 4 C 6, 3 80

7. C 6, 310

2!

8. C 2,1 C 5, 2 C 6, 2 C 2,1 C 2, 2 C 6, 2 270 9. C 4, 4 C 4, 3 C 4, 2 C 12, 4 11,880 10. a) C(11, 3) = 165 b) C(5, 3)C(6, 4) = 150 c) C(11,6) – C(2, 2)C(9, 4) = 336 d) C(5, 4)C(6, 3) – C(2, 2)C(3, 2)C(6, 3) = 40 11. a) C(7, 2) = 21 b) C(7, 3) = 35 12. C(6, 1) + C(6, 2) + C(6, 3) + C(6, 4) + C(6, 5) + C(6, 6) = 63 13. C 4, 3 C 4, 3 C 44, 7 2C 4, 4 C 4, 3 C 44, 6 C 4, 4 C 4, 4 C 44, 5 14. a) C 6, 6 1 b) C(6, 2) = 15 c) C(6, 4) + C(6, 5) + C(6, 6) = 22 15. C(3, 2) + C(3, 2) = 6 16. C(9, 3) = 84 17. C(13, 5)C(13, 8) = 1,656,369

65

1.5 – Combinations – Problem Set – SOLUTIONS (continued) 18. C(13, 3)C(11, 3) = 47,190 19. C 20, 4 C 10, 3 C 17, 4 C 10, 3 295,800 20. C(10, 6) – C(2, 2)C(8, 4) = 140 21. C(13, 5) – C(11, 5) = 825 22. a) C(7, 4) = 35 b) C(7, 4) + C(7, 5) + C(7, 6) + C(7, 7) = 64 c) C(7, 1) + C(7, 2) + C(7, 3) + C(7, 4) = 98 23. a) C(13, 10)C(39, 3) = 2,613,754 b) C(4, 1)C(13, 11)C(39, 2) = 231,192 24. a) C(4, 2)C(5, 1) = 30 b) C(4, 3) + C(5, 3) = 14 25. C(7, 5) + C(7, 6) + C(7, 7) = 29

66

1.6 – Binomial and Multinomial Theorems – SOLUTIONS 1. 0 3 1 4 2 5 3 6 32

2. 1 2 3 4 5 711 1 2 1 3 1 4 1 5 1 20

3. 1 1 1 2 2 1 3 3 1 4 4 120

2 2 2 2

4. 0 1 2 3 4

4 3 2 1 04 4 4 4 42 2 2 2 2x x x x x0 1 2 3 4y y y y y

3 2

42 3 4

8x 24x 32x 16xy y y y

5. 0 1 2 33 2 1 0 3 2 2 33 3 3 3a b a b a b a b a 3a b 3ab b

0 1 2 3

6. 4 6 4 6102x y 3360x y

6

7. 3

25 1 10x3 x x

8. 2 10 2 10123x 2y 608,256x y

10

9. 8

4212 1x 4958 x

10. 3 11 314k 1 364k 364

11

11. 3 7 3 710a b 120a b 120

7

12. 1121 1 2048

13. 531 2 1 4 1 243

14. 3 1 2 23 2 3 2 2 2 1 2 1 2 23 3 3 33!a x c a x a c x c a x c a x1 1 1 11!1!1!

12 2 23 3ac x c

2 2

3 6 3 2 2 2 4 2 4 2 2 2a x c 3a x 3a c 3x c 6ax c 3ax 3ac 3x c

67

1.6 – Binomial and Multinomial Theorems – SOLUTIONS (continued)

15. a) 2 3 06! x yz w 602!1!3!0!

b) 3 3 0 0 1 1 1 36! 6!x y z w 20 x y z w 120 20 120 1403!3!0!0! 1!1!1!3!

c) 61 1 1 1` 4096

d) 6 4 1 9

844 1 3

16. 2 1 00 3 2 2 4 12 2 2 45! 5! 5!x x 2 x x 2 x x 2 210x 2102!0!3! 1!2!2! 0!4!1!

17. a) 2 32 4 35

x 2y 80x y 803

b) 5 2 1 6

62 1 1

c) 521 2 1 1

d) 1 42 2 45

x 2y 80x y4

18. a) 2 1 1 0 2 22 24! 4!x 2y z 24x yz x 2y z 24x yz 24 24 482!1!1! 0!2!2!

b) 4 3 1 6

153 1 2

c) 41 2 1 1 16

68

1.7 – Combinatorial Equations – SOLUTIONS

1. n n 1 n 2 !

n n 1n 2 !

2.

n n 1 n 2 n 3 !n! n n 1 n 2

n 3 ! n 3 !

3.

n n 1 n 2 n 3 n 4 ! n n 1 n 2 n 3n!n 4 !4! n 4 !4! 24

4.

n!n k 1 ! k 1 ! n! n 1 k !k! n

n 1 ! n k 1 ! k 1 ! n 1 ! k 1n 1 k !k!

5.

2n 3 n 2 n 1 !

n 10n 4n 1 !

2n 3 n 2 n 10n 4

2 2

2

n 5n 6 n 10n 42n 5n 2 0

2n 1 n 2 01 1n , 2 reject n 22 2

6.

22n 2 2n 1 2n !

3n 9n 302n !

2

2 2

2

2n 2 2n 1 3n 9n 30

4n 6n 2 3n 9n 30n 3n 28 0

n 7 n 4 0n 4, 7 reject 4 n 7

69

1.7 – Combinatorial Equations – SOLUTIONS

7.

2 2

2

n 4 n 5 !18

n 2 n 5 !

2 2

2

2

n 4 n 5 !18

n 2 n 5 !

n 4 18n 2

n 2 n 218

n 2n 2 18

n 16

8.

n! n! 14n 1 ! n 1 ! 15

n n 1 ! n! 14n 1 ! n 1 n! 15

n 14n 1 1515n 14n 14

n 14

9.

3n 1 3n 3n 1 !

123n 1 !

2

2

3n 1 3n 12

9n 3n 129n 3n 12 0

3 3n 4 n 1 04 4n ,1 reject n 1

3 3

70

1.7 – Combinatorial Equations – SOLUTIONS

10.

2n 1 2n ! 3n 3n 1 ! n 1 n!50

2n ! 3n 1 ! 2n!

n 12n 1 3n 502

n 15n 1 502

10n 2 n 1 1009n 99

n 11

11. n! 14n!

n 5 ! n 4 !

n n 1 n 2 n 3 n 4 14n n 1 n 2 n 3

n 4 14n 18

18 18! 1532 18 2 !2!

12.

68 n 2 !3n!n 5 ! n 6!

2

2

3n n 1 n 2 n 3 n 4 68 n 2 n 3 n 4 n 5

3n n 1 68 n 5

3n 3n 68n 3403n 71n 340 0

n 17 3n 20 020 20n 17, reject n 173 3

19 19! 17117 19 2 !2!

71

1.7 – Combinatorial Equations – SOLUTIONS (continued)

13.

n 2 !4n! 97n 2 !2! n 4 !2!

2 2

2

4n n 1 n 2 n 397

2 24n n 1 n 2 n 3 194

4n 4n n 5n 6 1943n n 200 0

n 8 3n 25 025 25n 8, reject n 83 3

8!P 8, 2 568 2 !

14.

2 n 1 !5n!

n 3 ! n 1 4 !

2

2

2 n 1 !5n!n 3 ! n 5 !

5n n 1 n 2 2 n 1 n 2 n 3 n 4

5n 2 n 3 n 4

5n 2n 14n 240 2n 19n 240 2n 3 n 8

3 3n , 8 reject n 82 2

8 8! 563 8 3 !3!

72

1.7 – Combinatorial Equations – SOLUTIONS (continued)

15. n! 3024

n 4 !

2 2

4 3 2

4 3 2

2

n n 1 n 2 n 3 3024

n n n 5n 6 3024

n 6n 11n 6n 3024n 6n 11n 6n 3024 0

n 9 n 6 n 3n 56 0

n 6, 9 reject 6 n 912 12! 2209 12 9 !9!

16.

n 1 !n! 4n 2 !2! n 3 !2!

2 2

2

n n 1 n 1 n 24

2 2n n 1 n 1 n 2 8

n n n 3n 2 82n 4n 6 0

2 n 3 n 1 0n 3, 1 reject 1 n 3

6!P 6, 3 1206 3 !

73

1.7 – Combinatorial Equations – SOLUTIONS (continued)

17.

n 1 ! n!n 4 !n 1 3 !

2

2

n 1 ! n!n 2 ! n 4 !

n 1 n n 1 n n 1 n 2 n 3

n 1 n 2 n 3

n 1 n 5n 60 n 6n 50 n 1 n 5n 1, 5 reject n 1 n 5

7 7! 215 7 5 !5!

18. n! n! n! 5n

n 1 !1! n 2 !2! n 3 !3!

2 3 2

3

3

n n 1 n n 1 n 2n 5n

2 66n 3n n 1 n n 1 n 2 30n

6n 3n 3n n 3n 2n 30nn 5n 30n

n 25n 0n n 5 n 5 0

n 0, 5, 5 reject 0, 5 n 55!P 5, 2 20

5 2 !

74

1.7 – Combinatorial Equations – SOLUTIONS (continued)

19. 10n! 2n! n!n 1 ! n 2 ! n 3 !

2

2

10n 2n n 1 n n 1 n 2

10 2 n 1 n 1 n 2

10 2n 2 n 3n 20 n 5n 60 n 6 n 1n 1, 6 reject 1 n 6

15 15! 45512 15 12 !12!

20.

n 2 !n 1 4

n 2 2 !2!

2

2

n 2 n 3n 1 4

22 n 1 n 2 n 3 8

2n 2 n 5n 6 8n 3n 4 0

n 4 n 1 0n 4, 1 reject 1 n 4

4!P 4, 4 244 4 !

75

1.8 – Chapter 1 Review – SOLUTIONS 1. a) combination b) permutation c) combination d) combination e) permutation f) combination g) permutation h) permutation 2. a) C(5, 2) = 10

b) C 6, 310

2

3. 3 11 314k 1 364k

11

4. 3 7 6 1 5 4 3 2 1 15,120 5. C 13, 5 C 13, 6 C 26, 2 717,759,900 6. a) 8! 40,320 b) 8 1 ! 5040

7. 6!2 1202!3!1!

8.

6 5 4 3 2 16! 603!2! 3 2 1 2 1

9.

2 n 1 ! n!108n 1 ! n 2 !

2 2

2

2 n 1 n 108 n n 1

2n 2n 108 n nn 3n 108 0n 12 n 9 0

n 12, 9 reject 12 n 911 11! 559 11 9 !9!

10. 21 20 9 9 8 7 6 5 57,153,600 11. C 3, 3 C 10, 7 C 3, 2 C 10, 8 255 12. n B F 135 8 127, n B 64, n F 75, n B F x n B F n B n F n B F 127 64 75 x x 12

76

13. 2 4 22 4 48! 2x y 3 15,120x y2!4!2!

14. 26 26 9 10 10 10 21 21 9 10 10 10 2,115,000

15. a) 12

662

b) 7 5

351 1

c)

6 435

1 1420

2

16. 5 2 2 5 100

17. n 2 !42

n!

2

2

n 2 n 1 n!42

n!n 3n 2 42

n 3n 40 0n 8 n 5 5

n 8, 5 reject 8 n 5

18. 7 7 7 4 7 7 252 19. 2 at a time (2)(6) = (3)(4)…(one repeated product)…C(5, 2) – 1 = 9 3 at a time (3)(4)(5) = (2)(6)(5), (2)(3)(4) = (4)(6), (2)(3)(5) = (5)(6)…(three repeated

products)…C(5, 3) – 3 = 7 4 at a time (2)(3)(4)(5) = (4)(5)(6)…(one repeat product)…C(5, 4) – 1 = 4 5 at a time…C(5, 5) = 1 9 + 7 + 4 + 1 = 21

20. 12 1 !19,958,400

2

77

21.

n 2 !n! 24n 2 ! n 2 2 !2!

2 2

2

n 2 n 1n n 1 24

22n n 1 n 2 n 1 48

2n 2n n 3n 2 48n 5n 50 0

n 10 n 5 0n 10, 5 reject 5 n 10

10 10! 452 10 2 !2!

22. a) 5 1 ! 24 b) 4! 24

23.

6 5 4 3 2 1

304 3 2 1

24. C(5, 2) + C(5, 3) + C(5, 4) +C(5, 5) = 26

25.

10 7 4 23 3 2 2

63002!2!

26. n P 14, n H 11, n T 11, n P H 8, n H T 6, n P T 7,

n P H T 5 n P H T 30 x

30 x 14 11 11 8 6 7 5 x 10

27. C(5, 4)C(4, 3)(4!)(3!) + C(4, 4)C(5, 3)(4!)(3!) = 4320 28. a) 7 6 5 4 3 2520 b) 4 5 4 3 3 720 c) 6 5 2 4 3 720 d) 4 3 3 2 2 3 4 2 3 1 216 29. 71 1 1 1 16,384 30. a) C(52, 13)

b) C 52,13 C 39, 13 C 26,13 C 13, 134!

31. 37 37 37 36 36 36 3997 32. C(3, 1)C(4, 1)C(5, 3) = 120 33. 5 4 5 4 3 5 4 3 2 5 4 3 2 1 320

78

34.

n 1 n n 1 n 2 !20 n 1

n 2 !

2

2

n 1 n n 1 20 n 1

n n 20n n 20 0

n 5 n 4 0n 5, 4 reject 5 n 4

35. n D 24, n M 22, n D M 15 x 12 24 22 15 x 43

36.

7 5 32 2 2

1053!

37. 5 0 4 1 3 2 2 3 1 45 5 5 5 5x 3y x 3y x 3y x 3y x 3y

0 1 2 3 4

0 5 5 4 3 2 2 3 4 55x 3y x 15x y 90x y 270x y 405xy 243y

5

38. 6! 603!2!1!

39.

30 n 1 !n!n 4 ! n 1 3 !3!

30 n 1 n 2 n 3n n 1 n 2 n 3

6n 5

5 5! 15 5 5 !5!

40. 2 2 1 1 2 26! a b c 2 360a b c 3602!2!1!1!

41. 9 1 !20,160

2

42. a) C 14, 4 1001 b) C(3, 1)C(5, 1)C(4, 1)C(2, 1) = 120 c) C(8, 1)C(6, 1) = 48 d) C(3, 1)C(5, 1)C(4, 1)C(2, 0) + C(3, 1)C(5, 1)C(4, 0)C(2, 1) + C(3, 1)C(5, 0)C(4, 1)C(2, 1) + C(3, 0)C(5, 1)C(4, 1)C(2, 1) = 154

43. 10

2525

79

44. n F S G 100, n F 52, n S 63, n G 25, n F S G 4, n F S 24, n S G 11, n F G x

n F S G n F n S n G n F S n S G n F G n F S G

100 52 63 25 24 11 x 4 x 9

45. 3!4! 144 46. P(7, 5) = 2520

47. 10 3 7

2315 0 5

48. C(5, 1)C(5, 1)C(8, 2) = 700

49. 4 48 4 48 4 48

108,3362 3 3 2 4 1

50. 11! 4,989,6002!2!2!1!1!1!1!1!

51. n M 7, n B 10, n C 10, n M B 3, n M C 4, n B C 5

n M B C 18 2 16, n M B C x

16 7 10 10 3 4 5 x x 1

52. a) 6 6 6 6 1296 b) 6 5 4 3 360 c) 6 6 6 6 5 5 5 5 671 d) 6 5 4 3 5 4 3 2 240

53.

n 3 ! n 3 n 2 n 1 n n 1 !

n 3 n 2 n 1 nn 1 ! n 1 !

54. a) C(20, 5) – C(12, 5) – C(12, 4)C(8, 1) = 10,752 b) C(12, 2)C(8, 3) + C(12, 3)C(8, 2) + C(12, 4)C(8, 1) + C(12, 5) = 14,608 c) C(12, 3)C(8, 2) +C(12, 4)C(8, 1) + C(12, 5) = 10,912 55. 3!4! 144

56. 2 3 2 3 2 352y 3z 10 4y 27z 1080y z

3

57. 3 3 2 4 4 2 5 5 2 6 6 2 50 58. a) C(2, 2)C(23, 4) = 8855 b) C(13, 3)C(12, 3) = 62,920

80

59. n K M R 44 5 39, n K 28, n M 20, n R x, n K M 10, n M R 8, n K R 11, n K M R 6

n K M R n K n M n R n K M n M R n K R

n K M R

39 28 20 x 10 8 11 6 x 14 14 11 8 6 1

60. 20

77,52013

61. C 13, 3 C 10, 2 12,870 62. a) 5 5 5 125

b) 5!P 5, 3 60

5 3 !

63. a) C(6, 3) = 20 b) C(5, 3) + C(6, 3) + C(4, 3) = 34 c) C(5, 1)C(6, 1)C(4, 1) = 120

64. a) 5! 102!3!

b) 1 1 3 1 1 3 65. 2 5 4 3 2 1 240 66. C(10, 2)(2) = 90 67. n C 36, n S 55, n C S x, n C S 71 n C S n C n S n C S 71 36 55 x x 20 68. C(10, 2)C(4, 2)C(6, 3) = 5400 69. 7 3 4 84

81

70.

2n !

84n2n 3 !

2

3 2

3 2

2

2n 2n 1 2n 2 84n

2n 4n 6n 2 84n

8n 12n 4n 84n8n 12n 80n 0

4n 2n 3n 20 0

4n 2n 5 n 4 05 5n 0, , 4 reject 0, n 4

2 24 1 5! 10

2 5 2 !2!

71. 2 3 1 2 4 1 2 5 1 21

72. 5 5 53! 7501 1 22!1!

73. 21 1 02 44! a b c d 12abc1!1!2!0!

74.

5 4 3 2 15! 60

2!0! 2 1 1

75. n! 153

n 2 !2!

2

n n 1153

2n n 1 306

n n 306 0n 18 n 17 0

n 18, 17 reject 17 n 1818!P 18, 3 4896

18 3 !

76. 53 3 3 15 12

2 3 10 7

8 2x y 8x y 14y565 y 2x y 32x x

77. 12! 3,326,4003!3!2!2!1!1!

82

78. C(7, 4)C(10, 4) = 7350 79. a) P 10, 2 90 b) C 8, 3 56 80. C(9, 2) – (4)(3) = 24 81. C(4, 3)C(4, 3)C(44, 7) + 2C(4, 3)C(4, 4)C(44, 6) + C(4, 4)C(4, 4)C(44, 5) 82. n B S 320 x, n B 85, n S 200, n B S 60 n B S n B n S n B S 320 x 85 200 60 x 95

83.

7 5 3 1 6 4 22 2 3 1 2 2 2

602! 2!

84. 11! 92,4003!3!3!2!

85. 6 5 4 2 1 1 3 1 3 2 1 2 4 34,560 86. a) C(16, 5) = 4368 b) C(36, 5) = 376,992 c) C(13, 2)C(13, 3) = 22,308 d) C(26, 4)C(26, 1) + C(26, 5) = 454,480 e) C(4, 3)C(48, 2) = 4512 f) C(3, 1)C(12, 1)C(36, 2) + C(3, 2)C(12, 2)C(36, 1) = 29,808

87.

8 7 6 5 4 3 2 18! 8! 67208 5 ! 3! 3 2 1

88. 4 1 !4! 144 89. a) 3 3 2 3 2 1 15 b) 3 3 3 3 3 3 39 90. n n 1 n 2 n 5n 11

2

2

n 3n 2 5n 11n 8n 9 0

n 9 n 1 0n 1, 9 reject 1 n 9

91. C(8, 2)C(12, 3) + C(8, 3)C(12, 2) + C(8, 4)C(12, 1) + C(8, 5)C(12, 0) = 10,572

83

92. n! 2n!

n 2 ! n 4 !4!

2

2

n n 1 n 2 ! 2n n 1 n 2 n 3 n 4 !n 2 ! n 4 !4!

2n n 1 n 2 n 3n n 1

24n 2 n 3

112

12 n 5n 60 n 5n 60 n 6 n 1n 6, 1 reject 1 n 6

93. 2

222

4! 23x 2162!2! x

94. C(6, 2)C(4, 1) + C(6, 1)C(4, 2) = 96 95. P 7, 4 P 7, 5 P 7, 6 P 7, 7 13,440 96. P(20, 2) = 380

97. 13! 10,810,8003!4!2!2!1!1!

98. 2 2 2 2 25 6 5 6 6 6 7 6 7 8 6 8 9 6 9 45

99. 1 9 9 4 1 9 4 9 1 153 100. n M 85, n E 70, n M E 50, n M E 150 x 150 x 85 70 50 x 45 a) 85 – 50 = 35 b) 45