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5-1

Chapter 5

Exercises 5.1

1. a. ′S = {5, 6, 7}

b. ∪S T = {1, 2, 3, 4, 5, 7}

c. ∩S T = {1, 3}

d. ′ ∩S T = {5, 7}

2. a. {4, 5}′ =S

b. ∪S T = {1, 2, 3, 5}

c. ∩ = ∅S T

d. {5}′ ∩ =S T

3. a. ∪R S = {a, b, c, d, e, f}

b. ∩R S = {c}

c. ∩ = ∅S T

4. a. { , }∪ =R S a b

b. { }∩ =R S a

c. { , }′ =T a c

d. { , , }′∪ =T S a b c

5. ∅ , {1}, {2}, {1, 2}

6. , {1}∅

7. a. ∩ =M F {all male college students who like football}

b. ′M = {all female college students}

c. ′ ′∩M F = {all female college students who don’t like football}

d. ∪M F = {all male college students or all college students who like football}

8. a. ′S = {all corporations with headquarters not in New York City}

b. ′T = {all publicly owned corporations}

c. ∩S T = {all privately owned corporations with headquarters in New York City}

d. ′∩S T = {all publicly owned corporations with headquarters in New York City}

9. a. S = {1983, 1984, 1987, 1999, 2003, 2006}

b. T = {1980, 1983, 1985, 1989, 1991, 1995, 1996, 1997, 1998, 1999, 2003}

c. ∩S T = {1983, 1999, 2003}

d. S ∪ T = {1980, 1983, 1984, 1985, 1987, 1989, 1991, 1995, 1996, 1997, 1998, 1999, 2003, 2006}

e. ′ ∩S T = {1980, 1985, 1989, 1991, 1995, 1996, 1997, 1998}

f. ′∩S T = {1984, 1987, 2006}

10. a. A = {1981, 1982, 1985, 1986, 1988, 1991, 1993, 1998, 2000, 2001, 2005, 2007}

b. B = {1981, 1990, 1994, 2000, 2001, 2002}

c. ∩A B = {1981, 2000, 2001}

d. ′ ∩A B = {1990, 1994, 2002}

e. ′∩A B = {1982, 1985, 1986, 1988, 1991, 1993, 1998, 2005, 2007}

11. From 1980 to 2007, during only three years did the Standard and Poor’s Index increase by 2% or more during the first 5 days and not increase by 16% or more for that year.

12. From 1980 to 2007, during only nine years did the Standard and Poor’s Index decrease during the first five days but increase for the year.

13. a. { , , , }∪ =R S a b c e

( )′∪R S = {d, f}

b. ∪ ∪R S T = {a, b, c, e, f}

c. ∩R S = {a, c} ( )∩ ∩ = ∩ ∩ = ∅R S T R S T

Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.

Chapter 5: Sets and Counting ISM: Finite Math

5-2

d. ′T = {a, b, c, d} ( )′ ′∩ ∩ = ∩ ∩R S T R S T = {a, c}

e. ′R = {d, e, f}; ∩S T = {e} ( ) { }′ ′∩ ∩ = ∩ ∩ =R S T R S T e

f. ∪S T = {a, c, e, f}

g. { , , , };∪ =R S a b c e

∪R T = {a, b, c, e, f} ( ) ( )∪ ∩ ∪R S R T = {a, b, c, e}

h. ∩ =R S {a, c}; ∩ = ∅R T ( ) ( ) { , }∩ ∪ ∩ =R S R T a c

i. { , , }; { , , , }′ ′= =R d e f T a b c d

{ }′ ′∩ =R T d

14. a. ∩R S = {3, 5} ( )∩ ∩ = ∩ ∩ = ∅R S T R S T

b. ′T = {1, 3, 5} ( )′ ′∩ ∩ = ∩ ∩R S T R S T = {3, 5}

c. {1, 2}; {1}′ ′= ∩ =S R S

( )′ ′∩ ∩ = ∩ ∩ = ∅R S T R S T

d. {2, 4}′ = =R T

{2, 4}′ ∩ = =R T T

e. {1, 3, 4, 5}∪ =R S

f. ′ ∪ =R R U = {1, 2, 3, 4, 5}

g. {4}∩ =S T

( ) {1, 2, 3, 5}′∩ =S T

h. {1, 2}; {1, 3, 5}′ ′= =S T

{1, 2, 3, 5}′ ′∪ =S T

15. ( )′ ′ =S S

16. ′∩ = ∅S S

17. ′∪ =S S U

18. ∩ ∅ = ∅S

19. ( )′ ′∩ ∩ = ∩ ∩ = ∩ ∅ = ∅T S T S T T S

20. ∪ ∅ =S S

21. {divisions that had increases in labor costs or total revenue} = ∪L T

22. {divisions that did not make a profit} = ′P

23. {divisions that made a profit despite an increase in labor costs} = ∩L P

24. {divisions that had an increase in labor costs and were either unprofitable or did not increase their total revenue}

( )′ ′= ∩ ∪L P T

25. {profitable divisions with increases in labor costs and total revenue} = P ∩ L ∩ T

26. {divisions that were unprofitable or did not have increases in either labor costs or total revenue}

( )′ ′= ∪ ∪P L T

27. {applicants who have not received speeding tickets} = ′S

28. {applicants who have caused accidents and been arrested for drunk driving} = A ∩ D

29. {applicants who have received speeding tickets, caused accidents, or were arrested for drunk driving} = ∪ ∪S A D

30. {applicants who have not been arrested for drunk driving but have received speeding tickets or have caused accidents}

( )′= ∩ ∪D S A

31. {applicants who have not both caused accidents and received speeding tickets but who have been arrested for drunk driving}

( )′= ∩ ∩A S D

32. {applicants who have not caused accidents or have not been arrested for drunk driving} =

′ ′∪A D

33. ∩A D = {male students at Mount College}

34. ∩B C = {female teachers at Mount College}

35. ∩A B = {people who are both teachers and students at Mount College}


ISM: Finite Math Chapter 5: Sets and Counting

5-3

36. ∪B C = {teachers or females at Mount College}

37. ′∪ = ∪A C A D = {males or students at Mount College}

38. ( )′∩A D = {people at Mount College who are

not male students}

39. ′ =D C = {females at Mount College}

40. ∩ =D U D = {males at Mount College}

41. {people who don’t like strawberry ice cream} = ′S

42. {people who like vanilla but not chocolate ice cream} ′= ∩V C

43. {people who like vanilla or chocolate but not strawberry ice cream} = ( ) ′∪ ∩V C S

44. {people who don’t like any flavor of the three flavors of the ice cream} ( )′= ∪ ∪S V C

45. {people who like neither chocolate nor vanilla ice cream} = ( )′∪V C

46. {people who like only strawberry ice cream} = ′ ′∩ ∩S V C

47. a. R = {B, C, D, E}

b. S = {C, D, E, F}

c. T = {A, D, E, F}

d. ′R = {A, F} ′ ∪R S = {A, C, D, E, F}

e. ′ ∩R T = {A, F}

f. ∩R S = {C, D, E} ( ) { , }∩ ∩ = ∩ ∩ =R S T R S T D E

48. no toppings, butter, cheese, chives, butter and cheese, butter and chives, cheese and chives, all three toppings

49. Any subset with 2 as an element is an example. Possible answer: {2}

50. If T is a subset of S, then .∩ =S T T

51. If S is a subset of T, then .∪ =S T T

52. There are many possible subsets. Possible answer: R = {1}, S = {1, 2}, T = {2, 3}. Then

( ) {1, 2}∪ ∩ =R S T and ( ) {2}.∪ ∩ =R S T

53. True; 5 is an element of the set {3, 5, 7}.

54. True; {1, 3} is a subset of the set {1, 2, 3}.

55. True; {b} is a subset of the set {b, c}.

56. False; 0 is not an element of the set {1, 2, 3}.

57. False; 0 is not an element of the empty set ∅.

58. True; the empty set ∅ is a subset of any set.

59. True; any set is a subset of itself.

60. False; 1 is an element of the set {1}.

Exercises 5.2

1. ( ) ( ) ( ) ( )∪ = + − ∩n S T n S n T n S T

5 4 2 7= + − =

2. ( ) ( ) ( ) ( )17 13 921

∪ = + − ∩= + −=

n S T n S n T n S T

3. ( ) ( ) ( ) ( )∪ = + − ∩n S T n S n T n S T

15 = 7 + 8 – ( )∩n S T

( ) 7 8 15 0∩ = + − =n S T

4. ( ) ( ) ( ) ( )∪ = + − ∩n S T n S n T n S T

15 4 12 ( )= + − ∩n S T

( ) 4 12 15 1∩ = + − =n S T

5. ( ) ( ) ( ) ( )∪ = + − ∩n S T n S n T n S T

13 = n(S) + 7 – 5 n(S) = 13 – 7 + 5 = 11

6. ( ) ( ) ( ) ( )∪ = + − ∩n S T n S n T n S T

14 = 14 + n(T) – 6 n(T) = 14 – 14 + 6 = 6

7. S is a subset of T.

8. T is a subset of S.

9. Let P = {adults in South America fluent in Portuguese} and S = {adults in South America fluent in Spanish}. Then ∪P S = {adults in South America fluent in Portuguese or Spanish} and

∩P S = {adults in South America fluent in Portuguese and Spanish}. n(P) = 134, n(S) = 130, n(P ∪ S) = 245 (numbers in millions)



5-4

( ) ( ) ( ) ( )∪ = + − ∩n P S n P n S n P S

245 = 134 + 130 − n(P ∩ S) n(P ∩ S) = 134 + 130 − 245 = 19 19 million are fluent in both languages.

10. Let M = {fist-year students enrolled in a math course} and E = {first-year students enrolled in an English course}. Then ∪M E = {all first-year students} and

∩M E = {first-year students enrolled in both math and English}.

( ) 600, ( ) 400, ( ) 1000= ∩ = ∪ =n E n M E n M E

( ) ( ) ( ) ( )∪ = + − ∩n M E n M n E n M E

1000 ( ) 600 400= + −n M

n(M) = 1000 – 600 + 400 = 800 800 are taking a math course.

11. Let U = {all letters of the alphabet}, let V = {letters with vertical symmetry}, and H = {letters with horizontal symmetry}. Then ∪ =V H {letters with vertical or horizontal symmetry} and

∩ =V H {letters with both vertical and horizontal symmetry}. n(V) = 11, n(H) = 9, ( ) 4∩ =n V H

( ) ( ) ( ) ( )∪ = + − ∩n V H n V n H n V H

( ) 11 9 4 16∪ = + − =n V H

(( ) ) ( ) ( ) 26 16 10′∪ = − ∪ = − =n V H n U n V H

There are 10 letters with no symmetry.

12. Let N = {subscribers to Newsweek} and T = {subscribers to Time}. Then ∪N T = {subscribers to Newsweek or Time} and

∩N T = {subscribers to both Newsweek and Time}.

( ) 300, ( ) 200, ( ) 50= = ∩ =n N n T n N T

( ) ( ) ( ) ( )300 200 50450

∪ = + − ∩= + −=

n N T n N n T n N T

450 subscribe to at least one of these magazines.

13. Let A = {cars with automatic transmission} and P = {cars with power steering}. Then ∪A P = {cars with automatic transmission or power steering} and

∩A P = {cars with both automatic transmission and power steering}, n(A) = 325, n(P) = 216, ( ) 89∩ =n A P

( ) ( ) ( ) ( )325 216 89452

∪ = + − ∩= + −=

n A P n A n P n A P

452 cars were manufactured with at least one of the two options.

14. Let S = {investors in stocks} and B = {investors in bonds}. Then ∪S B = {investors in either stocks or bonds} and ∩S B = {investors in both stocks and bonds}.

( ) 80, ( ) 70, ( ) 100= = ∪ =n S n B n S B

( ) ( ) ( ) ( )∪ = + − ∩n S B n S n B n S B

100 80 70 ( )= + − ∩n S B

( ) 80 70 100 50∩ = + − =n S B

50 investors owned both.

15. Consists of points not in S but in T.

S T

16. Consists of points not in S and not in T.

S T

17. Consists of points in S or not in T.

S T

18. Consists of points not in S or not in T.

S T

19. ( )′ ′ ′∩ = ∪S T S T

Consists of points in S or not in T.

S T

20. ( )′ ′ ′∩ = ∪S T S T

Consists of points not in S or not in T.

S T



5-5

21. Consists of points in S but not in T or points in T but not in S.

S T

22. Consists of points in both S and T or points in neither S nor T.

S T

23. ( )∪ ∩ =S S T S

Consists of points in S.

S T

24. ( )′ ′∪ ∪ = ∪S T S S T

Consists of points in S or not in T.

S T

25. ′∪ =S S U Consists of all points.

S T

26. ′∩ = ∅S S Consists of no points.

S T

27. Consists of points in R and S but not in T.

R S

T

28. Consists of points in T but not in R and not in S.

R S

T

29. Consists of points in R or points in both S and T.

R S

T

30. Consists of points in both R and S or points in both R and T.

R S

T

31. Consists of points in R but not in S or points in both R and T.

R S

T

32. Consists of points not in R or points in S but not in T.

R S

T

33. Consists of points in both R and T.

R S

T

34. Consists of points in S but not in T.

R S

T

35. Consists of points not in R, S, or T.

R S

T

36. ( )′ ′ ′ ′∪ ∪ = ∩ ∩R S T R S T

Consists of points not in R, S, or T.

R S

T

37. Consists of points in R and T or points in S but not T.

R S

T

38. Consists of points in R or points not in S or T.

R S

T

39. ( )′ ′ ′ ′ ′ ′ ′∪ ∩ = ∪ ∪ = ∪S S T S S T S T

40. ( ) ( )′ ′ ′ ′ ′∩ ∪ = ∩ ∩ = ∩ ∩ = ∅T S T T S T T T S



5-6

41. ( )′ ′ ′∪ = ∩S T S T

42. ( )′ ′ ′∩ = ∪S T S T

43. ( )( )

′ ′ ′∪ ∩ = ∪ ∪′ ′= ∪ ∪

=

T S T T S TT T S

U

44. ( )′ ′ ′ ′∩ ∪ = ∪ ∪ = ∪S T S S T S S T

45. ′S

46. ( ) ( )′ ′∩ ∪ ∩S T S T

47. R ∩ Τ

48. ′ ′∩ ∩R S T

49. ′ ∩ ∩R S T

50. ( )′∩ ∪R T S

51.

R S

T

( )′∪ ∩T R S

52.

R S

T

( ) ( )′∩ ∪ ∩R T S T

53. First draw a Venn diagram for ( ) ( ) ( )′ ′ ′∩ ∪ ∩ ∪ ∩R S S T T R .

R S

T

The set consists of the complement. ( ) ( )′ ′ ′∩ ∩ ∪ ∩ ∩R S T R S T

54. Children from 5 to 18 years old who are legal aliens and employed

55. People who are not illegal aliens or everyone over the age of 18 who is employed

56. All citizens who are over the age of 18 or employed

57. Everyone over the age of 18 who is unemployed

58. No one, since there are no people who are citizens and legal aliens ( )∩ = ∅A B .

59. B is a subset of ′A , so .′ ′∪ =A B A Noncitizens who are unemployed

Exercises 5.3

1. 5 + 6 = 11

2. 15 + 20 = 35

3. 6 + 5 + 15 + 20 = 46

4. 14 + 19 + 10 + 11 = 54

5. 11

6. 19

7. 19 + 10 + 5 + 6 + 15 + 20 = 75

8. 10 + 5 + 6 = 21

9. 10 + 5 + 15 = 30

10. 19 + 10 + 5 + 6 + 11 + 15 + 20 = 86

11. ( ) ( ) ( ) 5 2 3′∩ = − ∩ = − =n S T n S n S T

( ) ( ) ( ) 6 2 4′ ∩ = − ∩ = − =n S T n T n S T

( ) ( ) ( ) ( )∪ = + − ∩n S T n S n T n S T

= 5 + 6 – 2 = 9 ( ) ( ) ( ) 14 9 5′ ′∩ = − ∪ = − =n S T n U n S T

TS4

23

5

12. ( ) ( ) – ( ) 11 – 7 4′∩ = ∩ = =n S T n S n S T

( ) ( ) ( ) 7 – 7 0′ ∩ = − ∩ = =n S T n T n S T

( ) ( ) ( ) – ( ) 11 7 – 7 11∪ = + ∩ = + =n S T n S n T n S T

( ) ( ) – ( ) 20 –11 9′ ′∩ = ∪ = =n S T n U n S T

TS0

74

9



5-7

13. ( ) ( ) ( ) ( )∩ = + − ∪n S T n S n T n S T

= 12 + 14 – 18 = 8 ( ) ( ) ( ) 12 8 4′∩ = − ∩ = − =n S T n S n S T

( ) ( ) ( ) 14 8 6′ ∩ = − ∩ = − =n S T n T n S T

( ) ( ) ( ) 20 18 2′ ′∩ = − ∪ = − =n S T n U n S T

TS6

84

2

14. n(S) = n(S ∩ T) + n(S ∪ T) – n(T) = 5 + 10 – 7 = 8 ( ) ( ) ( ) 8 6 14′= + = + =n U n S n S

( ) ( ) – ( ) 8 – 5 3′∩ = ∩ = =n S T n S n S T

( ) ( ) – ( ) 7 – 5 2′ ∩ = ∩ = =n S T n T n S T

( ) ( ) – ( ) 14 –10 4′ ′∩ = ∪ = =n S T n U n S T

TS2

53

4

15. ( ) ( ) ( )′ ′∪ = − ∩n S T n U n S T 75 40 35= − =

( ) ( ) ( ) ( )∩ = + − ∪n S T n S n T n S T = 15 + 25 – 35 = 5

( ) ( ) ( ) 15 5 10′∩ = − ∩ = − =n S T n S n S T

( ) ( ) ( ) 25 5 20′ ∩ = − ∩ = − =n S T n T n S T

TS20

510

40

16. ( ) ( ) ( ) 9 13 22′= + = + =n U n S n S

n(S ∪ T) = n(S) + n(T) – n(S ∩ T) = 9 + 11 – 5 = 15

( ) ( ) – ( ) 9 – 5 4′∩ = ∩ = =n S T n S n S T

( ) ( ) – ( ) 11– 5 6′ ∩ = ∩ = =n S T n T n S T

( ) ( ) – ( ) 22 –15 7′ ′∩ = ∪ = =n S T n U n S T

TS6

54

7

17. ( ) ( ) ( ) ( )∩ = + − ∪n S T n S n T n S T = 3 + 4 – 6 = 1

( ) ( ) ( ) 9 1 10′ ′= ∪ + ∩ = + =n U n S T n S T

( ) ( ) ( ) 3 1 2′∩ = − ∩ = − =n S T n S n S T

( ) ( ) ( ) 4 1 3′ ∩ = − ∩ = − =n S T n T n S T

( ) ( ) ( ) 10 6 4′ ′∩ = − ∪ = − =n S T n U n S T



5-8

TS3

12

4

18. ( ) ( ) ( ) ( ) 8 9 14 3∩ = + − ∪ = + − =n S T n S n T n S T

( ) ( ) ( ) 8 3 5′∩ = − ∩ = − =n S T n S n S T

( ) ( ) ( ) 9 3 6′∩ = − ∩ = − =n T S n T n S T

( ) ( ) ( ) 15 14 1′ ′∩ = − ∪ = − =n S T n U n S T

TS6

35

1

19. ( ) ( ) ( )′∩ ∩ = ∩ − ∩ ∩n R S T n R S n R S T = 7 – 2 = 5

( ) ( ) ( )′∩ ∩ = ∩ − ∩ ∩n R S T n R T n R S T = 6 – 2 = 4

( ) ( ) ( )′∩ ∩ = ∩ − ∩ ∩n R S T n S T n R S T = 5 – 2 = 3

( )′ ′∩ ∩n R S T ( ) ( ) ( )′ ′= − ∩ ∩ − ∩ ∩n R n R S T n R S T ( )− ∩ ∩n R S T = 17 – 5 – 4 – 2 = 6

( ) ( ) – ( ) – ( ) – ( ) 17 – 5 – 3 – 2 7′ ′ ′ ′∩ ∩ = ∩ ∩ ∩ ∩ ∩ ∩ = =n R S T n S n R S T n R S T n R S T

( )′ ′∩ ∩n R S T ( ) ( ) ( )′ ′= − ∩ ∩ − ∩ ∩n T n R S T n R S T ( )− ∩ ∩n R S T = 17 – 4 – 3 – 2 = 8

( )∪ ∪n R S T = 6 + 7 + 8 + 5 + 4 + 3 + 2 = 35

( ) ( ) ( )′ ′ ′∩ ∩ = − ∪ ∪n R S T n U n R S T = 44 – 35 = 9

T

R4

6S

3

8

52

7

9

20. ( ) ( ) – ( ) 1–1 0′∩ ∩ = ∩ ∩ ∩ = =n R S T n R S n R S T

( ) ( ) – ( ) 5 –1 4′∩ ∩ = ∩ ∩ ∩ = =n R S T n R T n R S T

( ) ( ) – ( ) 4 –1 3′∩ ∩ = ∩ ∩ ∩ = =n R S T n S T n R S T

( ) ( ) – ( ) – ( ) – ( ) 10 – 0 – 4 –1 5′ ′ ′ ′∩ ∩ = ∩ ∩ ∩ ∩ ∩ ∩ = =n R S T n R n R S T n R S T n R S T

( ) ( ) – ( ) – ( ) – ( ) 12 – 0 – 3 –1 8′ ′ ′ ′∩ ∩ = ∩ ∩ ∩ ∩ ∩ ∩ = =n R S T n S n R S T n R S T n R S T

( ) ( ) – ( ) – ( ) ( ) 10 – 4 – 3 –1 2′ ′ ′ ′∩ ∩ = ∩ ∩ ∩ ∩ = ∩ ∩ = =n R S T n T n R S T n R S T n R S T

( ) 5 8 2 0 4 3 1 23∪ ∪ = + + + + + + =n R S T

( ) ( ) – ( ) 29 – 23 6′ ′ ′∩ ∩ = ∪ ∪ = =n R S T n U n R S T

T

R

45

S

3

2

01 8

6

21. ( ) ( ) ( ) ( ) 21 7 14 14= ∪ + ∩ − = + − =n R n R S n R S n S

( ) ( ) ( ) 22 14 36′= + = + =n U n R n R

( ) ( ) ( )′∩ ∩ = ∩ − ∩ ∩n R S T n R S n R S T = 7 – 5 = 2

( )′∩ ∩n R S T = ( ) ( ) 11 5 6∩ − ∩ ∩ = − =n R T n R S T

( ) ( ) ( )′∩ ∩ = ∩ − ∩ ∩n R S T n S T n R S T = 9 – 5 = 4



5-9

( )′ ′∩ ∩n R S T ( ) ( ) ( )′ ′= − ∩ ∩ − ∩ ∩n R n R S T n R S T ( )− ∩ ∩n R S T = 14 – 2 – 6 – 5 = 1

( )′ ′∩ ∩n R S T ( ) ( ) ( )′ ′= − ∩ ∩ − ∩ ∩n S n R S T n R S T – ( )∩ ∩n R S T = 14 – 2 – 4 – 5 = 3

( )′ ′∩ ∩n R S T ( ) ( ) ( )′ ′= − ∩ ∩ − ∩ ∩n T n R S T n R S T – ( )∩ ∩n R S T = 22 – 6 – 4 – 5 = 7

( ) 1 3 7 2 6 4 5 28∪ ∪ = + + + + + + =n R S T

( ) ( ) ( )′ ′ ′∩ ∩ = − ∪ ∪n R S T n U n R S T = 36 – 28 = 8

TR

4

6

S 3

75

21

8

22. n(R ∪ T) = n(R) + n(T) – n(R ∩ T) = 22 + 26 – 8 = 40 n((R ∪ T) ∩ S) = n((R ∩ S) ∪ (T ∩ S)) = n(R ∩ S) + n(T ∩ S) – n(R ∩ S ∩ T) = 4 + 6 – 1 = 9 n(S) = n(R ∪ S ∪ T) + n((R ∪ T) ∩ S) – n(R ∪ T) = 45 + 9 – 40 = 14

( ) ( ) – ( ) 4 –1 3′∩ ∩ = ∩ ∩ ∩ = =n R S T n R S n R S T

( ) ( ) – ( ) 8 –1 7′∩ ∩ = ∩ ∩ ∩ = =n R S T n R T n R S T

( ) ( ) – ( ) 6 –1 5′∩ ∩ = ∩ ∩ ∩ = =n R S T n S T n R S T

( ) ( ) – ( ) – ( ) – ( ) 22 – 3 – 7 –1 11′ ′ ′ ′∩ ∩ = ∩ ∩ ∩ ∩ ∩ ∩ = =n R S T n R n R S T n R S T n R S T

( ) ( ) – ( ) – ( ) – ( ) 14 – 3 – 5 –1 5′ ′ ′ ′∩ ∩ = ∩ ∩ ∩ ∩ ∩ ∩ = =n R S T n S n R S T n R S T n R S T

( ) ( ) – ( ) – ( ) – ( ) 26 – 7 – 5 –1 13′ ′ ′ ′∩ ∩ = ∩ ∩ ∩ ∩ ∩ ∩ = =n R S T n T n R S T n R S T n R S T

( ) ( ) – ( ) 64 – 45 19′ ′ ′∩ ∩ = ∪ ∪ = =n R S T n U n R S T

T

R

711

S

513

31 5

19

23. Let U = {high school students surveyed}, R = {students who like rock music}, and H = {students who like hip-hop music}. n(U) = 70; n(R) = 35; n(H) = 15; n(R ∩ H) = 5

( ) ( ) ( ) ( )∪ = + − ∩n R H n R n H n R H = 35 + 15 – 5 = 45

(( ) ) ( ) ( ) 70 45 25′∪ = − ∪ = − =n R H n U n R H

25 students do not like either rock or hip-hop music.

24. Let U = {Nobel Prizes awarded}, L = {Nobel Prizes in literature}, and S = {Nobel Prizes awarded to Scandinavians}. n(U) = 797; n(L) = 101; n(S) = 55; n(L ∩ S) = 15 The number of Literature or Scandinavian winners = no. literature + no. Scandinavians – no. Literature and Scandinavians = 101 + 55 – 15 = 141. Since total number of Nobel prizes is 797, then the number of non-literature Nobel prizes awarded to non-Scandinavians is 797 − 141 = 656.

25. Let U = {lines}, V = {lines with verbs}, A = {lines with adjectives} n(U) = 14; n(V) = 11; n(A) = 9;

( ) 7∩ =n V A

( ) ( ) ( ) 11 7 4′∩ = − ∩ = − =n V A n V n V A

Four lines have a verb with no adjective. ( ) ( ) ( ) 9 7 2′∩ = − ∩ = − =n V A n A n V A

Two lines have an adjective but no verb. ( ) ( ) ( ) ( )∪ = + − ∩n V A n V n A n V A = 11 + 9 – 7 = 13

(( ) ) ( ) ( )′∪ = − ∩n V A n U n V A = 14 – 13 = 1

One line has neither an adjective nor a verb.



5-10

For Exercises 26−30, let U = {students who took the exam}, F = {students who correctly answered the first question}, S = {students who correctly answered the second question}. Then n(U) = 130, n(F) = 90, n(S) = 62, ( ) 50.∩ =n F S Draw and complete the Venn

diagram as follows.

SF12

5040

28

U

26. n(F ∪ S) = 40 + 50 + 12 = 102

27. (( ) ) 28′∪ =n F S

28. (( ) ( )) 40 12 52′ ′∩ ∪ ∩ = + =n F S F S

29. ( ) 12′∩ =n S F

30. ( ) 28 40 68′ = + =n S

31. Let U = {students in finite math}, M = {male students}, B = {students who are business majors}, and F = {first-year students}. n(U) = 35; n(M) = 22; n(B) = 19; n(F) = 27;

( ) 14; ( ) 17;∩ = ∩ =n M B n M F

( ) 15; ( ) 11∩ = ∩ ∩ =n B F n M B F

a. Draw a Venn diagram as shown.

F

M

62

B

46

311

1

2

U

b. ( ) 2′ ′ ′∩ ∩ =n F M B

There are two upperclass women nonbusiness majors.

c. ( ) 1 4 5′ ∩ = + =n M B

There are five women business majors.

32. Let U = {college faculty}, J = {faculty who jog}, S = {faculty who swim}, and C = {faculty who cycle}. n(U) = 100; n(J) = 45; n(S) = 30; n(C) = 20; n(J ∩ S) = 6; n(J ∩ C) = 1; n(S ∩ C) = 5; n(J ∩ S ∩ C) = 1 Draw a Venn diagram as shown.

C

J

039

S

415

51 20

16

U

(( ) ) 16′∪ ∪ =n J S C

16 members do not do any of these three activities.

( ) 39′ ′∩ ∩ =n J S C

39 members just jog.

For Exercises 33–38, let U = {people surveyed}, I = {people who learned from the Internet}, T = {people who learned from television}, N = {people who learned from newspapers}. Then n(U) = 400, n(I) = 180, n(T) = 190, n(N) = 190,

( ) 80, ( ) 90,∩ = ∩ =n I T n I N

( ) 50, ( ) 30.∩ = ∩ ∩ =n T N n I T N Draw and complete

the Venn diagram as follows.

N

I

60

40T

2080

5030

90

30

U

33. (( ) ( )) 40 50 80 20 190′ ′∩ ∪ ∩ = + + + =n I N I N

34. ( ) 80′ ′∩ ∩ =n I T N

35. (( ) ) 40 90 50 180′∪ ∩ = + + =n I T N

36. n((I ∩ T) ∪ (I ∩ N) ∪ (T ∩ N)) = 50 + 60 + 20 + 30 = 160

37. (( ) ( ) ( ))40 90 80210

′ ′ ′ ′ ′ ′∩ ∩ ∪ ∩ ∩ ∪ ∩ ∩= + +=

n I T N I T N I T N

38. ( ) 50′∩ ∩ =n I T N

39. Draw and complete the Venn diagram as follows:

B

C5

40P

1015

150

15U

40 + 15 +15 + 15 + 5 + 10 + 0 = 100 The correct answer is (d).



5-11

For Exercises 40–44, n(U) = 4000, n(F) = 2000, n(S) = 3000, n(L) = 500,

( ) 1500, ( ) 300,∩ = ∩ =n F S n F L

( ) 200, ( ) 50.∩ = ∩ ∩ =n S L n F S L Draw and complete

the Venn diagram as follows.

L

F

250

250

S

15050

1450

501350

450

U

40. ( ) 250 150 50 450∩ ∪ = + + =L F S

41. ( ) 450′∪ ∪ =L F S

42. 4000 – 500 3500′ = =L

43. 4000 250 3750′∪ ∪ = − =L S F

44. 250′ ′∩ ∩ =F S L

45. Let U = {college students surveyed}, F = {first-year students}, D = {voted Democratic}, n(U) = 100; n(F) = 50; n(D) = 55

( ) (( ) ) 25′ ′ ′∩ = ∪ =n F D n F D

( ) ( ) (( ) )′∪ = − ∪n F D n U n F D 100 25 75= − =

( ) ( ) ( ) ( )50 55 7530

∩ = + − ∪= + −=

n F D n F n D n F D

30 freshmen voted Democratic.

46. Let U = {workers}, G = {college graduates}, M = {union members}. n(U) = 100; ( ) 60; ( ) 20;′ ′= ∩ =n G n G M

n(M) = 30 ( ) ( ) ( ) 20 30 50′∪ = ∩ + = + =n G M n G M n M

( ) (( ) )( ) – (( ))

100 – 5050

′ ′ ′∩ = ∪= ∪==

n G M n G Mn U n G M

50 workers were neither college graduates nor union members.

47. Let U = {students}, D = {students who passed the diagnostic test}, C = {students who passed the course}. n(U) = 30; n(D) = 21; n(C) = 23; ( ) 2′∩ =n D C

( ) ( ) ( ) 21 2 19′∩ = − ∩ = − =n D C n D n D C

( ) ( ) ( )′ ∩ = − ∩n D C n C n D C = 23 – 19 = 4

Four students passed the course even though they failed the diagnostic test.

48. Let U = {applicants}, P = {pilots}, V = {veterans}. n(P) = 35; n(V) = 20;

( ) 30; ( ) 50′ ′ ′∩ = ∩ =n P V n P V

( ) ( ) ( ) 30 20 50′∪ = ∩ + = + =n P V n P V n V

( ) ( ) (( ) )( ) ( )

50 50100

′= ∪ + ∪′ ′= ∪ + ∩

= +=

n U n P V n P Vn P V n P V

There were 100 applicants.

For Exercises 49−54, let U = {students}, M = {male students}, B = {biology majors}. Then n(U) = 52, n(M ∩ B) = 5, ( ) 15,′ ∩ =n M B and

( ) 12.′ ′∩ =n M B Therefore n(B) = 5 + 15 = 20,

n(M ∪ B) = 52 − 12 = 40, so 40 = n(M) + 20 − 5 (inclusion-exclusion) and n(M) = 40 − 20 + 5 = 25. Draw and complete the Venn diagram as follows.

BM15

520

12

U

49. 40

50. 25

51. 27

52. 20

53. 20

54. 32

For Exercises 55–62, let U = {surveyed students}, R = {students who like rock}, C = {students who like country}, J = {students who like jazz}. Then n(U) = 190, n(R) = 114, n(C) = 50, n(R ∩ J) = 15, ( ) 11,∩ =n C J

( ) 20,′ ′∩ ∩ =n R C J ( ) 10,′∩ ∩ =n R C J

( ) 9′∩ ∩ =n R C J and (( ) ) 20.′∪ ∪ =n R C J Draw

and complete the Venn diagram as follows.

J

R

10

90C

620

95

30

20

U

55. ( ) 90′ ′∩ ∩ =n R C J

56. ( ) 30 6 36′ ∩ = + =n R C



5-12

57. ( ) 6′ ∩ ∩ =n R C J

58. (( ) ) 30 20 6 56′∪ ∩ = + + =n C J R

59. (( ) ( ) ( )) 90 30 20 140n R C J R C J R C J′ ′ ′ ′ ′ ′∩ ∩ ∪ ∩ ∩ ∪ ∩ ∩ = + + =

60. n(R ∩ C ∩ J) = 5

61. (( ) ( ) ( )) 9 10 6 530

n R C R J C J∩ ∪ ∩ ∪ ∩ = + + +=

62. (( ) ) 20 20 40′∪ = + =n R C

63. Let U = {executives}, F = {executives who read Fortune}, T = {executives who read Time}, M = {executives who read Money}. n(U) = 180, n(F) = 75, n(T) = 70, n(M) = 55, ( ) 25, ( ) 25, ( ) 5n F T n T M n M T N∩ = ∩ = ∩ ∩ =

Draw and complete the Venn diagram as follows.

M

F

5

45T

2025

205

25

35

U

(( ) ) 35n F T M ′∩ ∩ =

64. Let U = {small businesses which had failed}, R = {undercapitalized}, S = {inexperienced management}, T = {poor location}. n(U) = 100, n(R ∪ S ∪ T) = 95, n(R ∩ S ∩ T) = 4, ( ) 40, ( ) 15,′ ′ ′ ′∩ ∩ = ∩ ∩ =n R S T n R S T n(R ∩ S) = 7,

n(R ∩ T) = 9, n(S ∩ T) = 10 Draw and complete the Venn diagram as follows.

T

R

540

S

622

34

15

5

U

n(T) = 22 + 5 + 6 + 4 = 37 37 businesses had poor location. n(R) = 40 + 3 + 5 + 4 = 52 n(S) = 15 + 3 + 6 + 4 = 28 The most prevalent characteristic was being undercapitalized.

65. Let U = {students}, P = {students who play piano}, V = {students who play violin} and C = {students who play clarinet}. Then let x = n(P ∩ V ∩ C) and complete the Venn diagram as follows.

C

P

8 – x

37 + x

V

10 – x

10 + x

20 – xx

12 + x

0

U

37 + x + 12 + x + 10 + x + (20 − x) + (8 − x) + (10 − x) + x = 97 + x = 100, so x = 3.



5-13

Exercises 5.4

1. 3 · 5 = 15 routes

2. 2 · 3 = 6 outfits

3. 26 · 26 = 676 words

4. 26 · 25 = 650 words

5. 20 · 19 = 380 tickets

6. 1

(20 19) 1902

⋅ = tickets

7. 5 · 4 = 20 ways

8. 26 · 26 · 10 · 10 · 10 · 10 = 6,760,000 license plates

9. 5 · 4 · 3 · 2 · 1 = 120 ways

10. 4 · 3 · 2 · 1 = 24 ways

11. 2 · 2 · 2 · 2 · 2 · 2 = 62 64= sequences

12. There will be only one choice for the operation 6.

52 2 2 2 2 1 2 1 32⋅ ⋅ ⋅ ⋅ ⋅ = ⋅ = sequences

13. 20 · 19 · 18 = 6840 possibilities

14. 8 · 7 · 6 · 5 · 4 · 3 · 2 · 1 = 40,320 orderings

15. 30 · 29 = 870 ways

16. 2 · 3 · 2 = 12 outfits

17. 4 · 3 · 2 · 1 = 24 words

18. 3 · 2 = 6 ways

19. 52 2 2 2 2 2 32⋅ ⋅ ⋅ ⋅ = = ways

20. The number of distinct pairs of initials is 26 · 26 = 676. Since there are 700 employees and only 676 possible pairs of initials, at least two have the same pair.

21. 3 · 12 · 10 · 10 · 10 · 10 = 360,000 serial numbers

22. The unique letters are s, t, a, i, and c. 5 · 4 · 3 · 2 = 120

23. a. 7 · 7 · 7 · 7 = 2401 words

b. 7 · 6 · 5 · 4 = 840 words

c. 1 · 7 · 7 · 7 = 343 words

d. Let operation 1 be choosing the letter at the end. 2 · 6 · 5 · 4 = 240 words

24. a. 8 · 7 · 6 · 5 · 4 · 3 · 2 · 1 = 40,320 ways

b. Front row can be arranged 3 · 2 · 1 = 6 ways. Back row can be arranged 5 · 4 · 3 · 2 · 1 = 120 ways. Thus, they can be arranged 6 · 120 = 720 ways.

25. a. 9 · 8 · 7 · 6 · 5 · 4 · 3 · 2 · 1 = 362,880 batting orders

b. 8 · 7 · 6 · 5 · 4 · 3 · 2 · 1 · 1 = 40,320 batting orders

c. 1 · 6 · 5 · 4 · 3 · 2 · 1 · 1 · 1 = 720 batting orders

26. A Venn diagram with two circles has 4 regions that can be shaded. Let each region be an operation. Each operation can be performed two ways, shaded or not shaded. 2 · 2 · 2 · 2 = 16 ways

27. A Venn diagram with three circles has 8 regions that can be shaded. Let each region be an operation. Each operation can be performed two ways, shaded or not shaded. 2 · 2 · 2 · 2 · 2 · 2 · 2 · 2 = 256 ways

28. Let n be the number of members of the club. Then n(n – 1) = 506. Solve the equation for n:

2( 1) 506

– – 506 0( 23)( 22) 0

23, 22

− ==

− + == −

n n

n nn n

n

Thus, the club has 23 members.

29. Each statement has three options. 63 3 3 3 3 3 3 729⋅ ⋅ ⋅ ⋅ ⋅ = = ways

30. She has 4 · 3 = 12 different combinations. Thus, she needs to recruit 12 · 10 = 120 volunteers.

31. 6 · 7 · 4 = 168 days



5-14

32. The student can enter 7 different ways and exit 6 different ways. The total number of ways to enter and exit is therefore 7 ⋅ 6 = 42.

33. The sequence is DLLLDDD. The number of

license plates is 3 3(10)(26) (10) 175,760,000.=

34. 3 · 6 = 18

35. The number of different area codes is (8)(2)(10) = 160 area codes.

36. The number of different area codes is (8)(10)(10) = 800

37. Let the winning team (I) consist of the players A,B,C,D,E,F,G,H, I, and J. Then let AB represent player A shaking hands with player B. Then the number of ways all that player A can shake hands is AB, AC, AD, AE, AF, AG, AH, AI, AJ = 9. Player B: BC, BD, BE, BF, BG, BH, BI, BJ = 8 ways (BA is the same handshake as AB and players cannot shake hands with themselves so that BB is not a valid handshake). Player C can shake hands 7 different ways , 6 handshakes for player D, 5 for player E, and so on. So the total number of handshakes with all the players from the winning team is 9 + 8 + 7 + 6 + 5 + 4 + 3 + 2 + 1 = 45 = (10(9))/2 = 45. Now, represent the players from team II (losing team) as K, L, M, N, O, P, Q, R, S, T. Then A has 10 handshakes: AK, AL, AM, AN, AO, AP, AQ, AR, AS, AT. So does B, C, D, ... J. So the number of different handshakes between the winning team members and losing team members is 10 10 10 10 10 10 10 10 10 10

100.+ + + + + + + + +

=

So the total number of handshakes is 100 + 45 = 145 .

38. 3 · 20 = 60 chairs

39. The first house has four choices of color. Each house after that has only three choices. 4 · 3 · 3 · 3 · 3 · 3 = 972 ways

40. The last digit must be odd. 7 · 7 · 4 = 196

41. 104 4 4 4 4 4 4 4 4 4 41,048,576 ways

⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ==

42. 10 · 9 = 90 ways

43. 4 · 4 = 16 ways

44. 6 · 10 · 5 = 300 ways

45. The first song can be any one of the three waltzes, the second song any of the three tangos, the third any of the two remaining waltzes, and so on. Altogether, there are 3 · 3 · 2 · 2 · 1 · 1 = 36 different arrangements of the three waltzes and three tangos. The correct answer is (d).

46. Each bottle has five possible locations to be delivered.

105 5 5 5 5 5 5 5 5 5 5 9,765,625 ways⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ = =

47. 4 · 2 · 3 = 24 computers

48. A voter has 7 choices for mayor, 4 choices for city council president and 2 choices (yes or no) for each of the 12 propositions. Therefore

127 4 2 114,688⋅ ⋅ = different ballots may be cast. If a voter can leave any item blank then there are 8 choices for mayor, 5 for city council president and 3 (yes, no or blank) for each of the 12 propositions, which gives

128 5 3 21,257,640⋅ ⋅ = different ballots.

49. Let operation 1 be choosing a box. There are 5 choices. Let operation 2 be choosing a wrapping paper. There are 11 choices (including not choosing a wrapping paper). Let operation 3 be choosing a ribbon with a special item for the bow. There are 7 · 2 · 10 + 1 = 141 choices (including not choosing a special item and not choosing a ribbon). There are 5 · 11 · 141 = 7755 ways to gift wrap.

50. 5 · 5 · 2 = 50 bags

51. 3 · 5 + 3 · 6 = 33 segments

52. 5 · 2 · 3 · 2 · 2 · 8 = 960 cars

53. 6 ⋅ 9 ⋅ 6 = 324

54. James has 100 choices. Janet has 14 choices. Sandy has 4 choices. There are 100 · 14 · 4 = 5600 triples.

55. 7 · 6 · 6 · 6 = 1512 numbers

56. A basket can have 0, 1, 2, 3, or 4 apples, so there are 5 possibilities for apples. Likewise, there are 4 possibilities for pears and 5 possibilities for oranges. Including the possibility of no fruit in a basket, there are 5 · 4 · 5 = 100 baskets. Thus there are 100 – 1 = 99 baskets if the basket must contain at least one piece of fruit.



5-15

57. 910 –1 999,999,999= numbers

58. 8 · 1 · 6 · 1 · 4 · 1 · 2 · 1 = 384 ways

59. 2 · 2 · 2 · 2 = 16 paths

60. Let n be the number of each article of clothing. The possible number of combinations is

3⋅ ⋅ =n n n n . We want 3 365.>n This occurs when n > 7. Thus she needs 8 of each article of clothing.

61. 6 · 1 · 4 · 1 · 2 · 1 = 48 ways

62. 26 · 25 · 25 = 16,250 words.

63. The proponents have a choice to start with the leftmost seat or to start with the rightmost seat. There are 4 · 3 · 2 · 1 = 24 ways to order the proponents. Thus the proponents can be seated 2 · 24 = 48 ways. Once the proponents have been seated there are 4 · 3 · 2 · 1 = 24 ways to order the opponents. Thus all 8 panelists can be seated in 48 · 24 = 1152 ways.

64. There are 1 ⋅ 26 ⋅ 26 = 676 three-letter call letters and 1 ⋅ 26 ⋅ 26 ⋅ 26 = 17,576 four-letter call letters; 676 + 17,576 = 18,252.

65. A five-digit number must begin with the digits 1−9; the next two digits can be chosen freely and the final two are determined by matching the first two. Hence there are 9 ⋅ 10 ⋅ 10 ⋅ 1 ⋅ 1 = 900 five-digit palindromes.

66. A six-digit number must begin with the digits 1−9; the next two digits can be chosen freely and the final three are determined by matching the first three. Hence there are 9 ⋅ 10 ⋅ 10 ⋅ 1 ⋅ 1 ⋅ 1 = 900 six-digit palindromes.

67. A restaurant has 5 appetizers and 6 salads on its menu. How many ways can a diner select an appetizer and salad combination?

Exercises 5.5

1. P(4, 2) = 4 · 3 = 12

2. P(5, 1) = 5 = 5

3. P(6, 3) = 6 · 5 · 4 = 120

4. P(5, 4) = 5 · 4 · 3 · 2 = 120

5. (10, 3) 10 9 8

(10, 3) 1203! 3 2 1

⋅ ⋅= = =⋅ ⋅

PC

6. (12, 2) 12 11

(12, 2) 662! 2 1

⋅= = =⋅

PC

7. (5, 4) 5 4 3 2

(5, 4) 54! 4 3 2 1

⋅ ⋅ ⋅= = =⋅ ⋅ ⋅

PC

8. (6, 3) 6 5 4

(6, 3) 203! 3 2 1

⋅ ⋅= = =⋅ ⋅

PC

9. P(7, 1) = 7

10. P(5, 5) = 5 · 4 · 3 · 2 · 1 = 120

11. P(n, 1) = n

12. 2( , 2) ( –1) –= ⋅ =P n n n n n

13. (4, 4) 4 3 2 1

(4, 4) 14! 4 3 2 1

⋅ ⋅ ⋅= = =⋅ ⋅ ⋅

PC

14. ( , 2) ( –1) ( 1)

( , 2)2! 2 1 2

⋅ −= = =⋅

P n n n n nC n

15. ( , – 2)

( , – 2)( – 2)!

( –1) 4 3

( – 2) ( – 3) 2 1( –1)

2 1( 1)

2

=

⋅ ⋅=⋅ ⋅

⋅=⋅−=

P n nC n n

nn n

n nn n

n n

16. ( , 1)

( , 1)1! 1

= = =P n nC n n

17. 6! = 6 · 5 · 4 · 3 · 2 · 1 = 720

18.

10! 10 9 8 7 6 5 4 3 2 1

4! 4 3 2 110 9 8 7 6 5151, 200

⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅=⋅ ⋅ ⋅

= ⋅ ⋅ ⋅ ⋅ ⋅=

19. 9! 9 8 7 6 5 4 3 2 1

9 8 727! 7 6 5 4 3 2 1

⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅= = ⋅ =⋅ ⋅ ⋅ ⋅ ⋅ ⋅

20. 7! = 7 · 6 · 5 · 4 · 3 · 2 · 1 = 5040

21. P(4, 4) = 4 · 3 · 2 · 1 = 24 ways



5-16

22. P(6, 2) = 6 · 5 = 30 ways

23. (9, 2) 9 8

(9, 2) 362! 2 1

⋅= = =⋅

PC selections

24. (5, 3) 5 4 3

(5, 3) 103! 3 2 1

⋅ ⋅= = =⋅ ⋅

PC pizzas

25. P(15, 3) = 15 · 14 · 13 = 2730 ways

26. P(6, 6) = 6 · 5 · 4 · 3 · 2 · 1 = 720 ways

27. (10, 4) 10 9 8 7

(10, 4)4! 4 3 2 1

⋅ ⋅ ⋅= =⋅ ⋅ ⋅

PC

= 210 deluxe chocolate banana splits

28. (10, 4) 10 9 8 7

(10, 4) 2104! 4 3 2 1

⋅ ⋅ ⋅= = =⋅ ⋅ ⋅

PC ways

29. (10, 6) 10 9 8 7 6 5

(10, 6)6! 6 5 4 3 2 1

⋅ ⋅ ⋅ ⋅ ⋅= =⋅ ⋅ ⋅ ⋅ ⋅

PC

= 210 ways

30. Selecting four out of 10 sweaters to take on a trip is the same as selecting six out of 10 sweaters to leave at home.

31. P(5, 5) = 5 · 4 · 3 · 2 · 1 = 120 ways

32. (9, 3) 9 8 7

(9, 3) 843! 3 2 1

⋅ ⋅= = =⋅ ⋅

PC ways

(Note that this is the same as computing C(9, 6).)

33. (8, 2) 8 7

(8, 2) 282! 2 1

⋅= = =⋅

PC games

34. (8, 5) 8 7 6 5 4

(8, 5) 565! 5 4 3 2 1

⋅ ⋅ ⋅ ⋅= = =⋅ ⋅ ⋅ ⋅

PC ways

35. P(10, 5) = 10 · 9 · 8 · 7 · 6 = 30,240 ways

36. (10, 5) 10 9 8 7 6

(10, 5) 2525! 5 4 3 2 1

⋅ ⋅ ⋅ ⋅= = =⋅ ⋅ ⋅ ⋅

PC ways

37. (10, 5) 10 9 8 7 6

(10, 5)5! 5 4 3 2 1

⋅ ⋅ ⋅ ⋅= =⋅ ⋅ ⋅ ⋅

PC

= 252 ways

38. P(10, 5) = 10 · 9 · 8 · 7 · 6 = 30,240 ways

39. P(35, 5) = 35 · 34 · 33 · 32 · 31 = 38,955,840 ways

40. (20, 4)

(20, 4)4!

20 19 18 17

4 3 2 14845 ways

=

⋅ ⋅ ⋅=⋅ ⋅ ⋅

=

PC

41. (100, 3) 100 99 98

(100, 3)3! 3 2 1

⋅ ⋅= =⋅ ⋅

PC

= 161,700 samples are possible (7, 3) 7 6 5

(7, 3)3! 3 2 1

⋅ ⋅= =⋅ ⋅

PC

= 35 samples consist of all defective diskettes

42. (7, 5)

(7, 5)5!

7 6 5 4 3

5 4 3 2 121 possibilities

=

⋅ ⋅ ⋅ ⋅=⋅ ⋅ ⋅ ⋅

=

PC

43. P(26, 3) = 26 · 25 · 24 = 15,600 words

44. (40, 5) 40 39 38 37 3678,960,960 possibilities

= ⋅ ⋅ ⋅ ⋅=

P

45. (100, 5)

(100, 5)5!

= PC

100 99 98 97 96

5 4 3 2 1

⋅ ⋅ ⋅ ⋅=⋅ ⋅ ⋅ ⋅

= 75,287,520 ways

46. (1000, 3) 1000 999 998997,002,000 possibilities

= ⋅ ⋅=

P

47. (52, 5) 52 51 50 49 48

(52, 5)5! 5 4 3 2 1

⋅ ⋅ ⋅ ⋅= =⋅ ⋅ ⋅ ⋅

PC

= 2,598,960 hands

48. There are 8 aces or kings in a deck. (8, 5) 8 7 6 5 4

(8, 5)5! 5 4 3 2 1

⋅ ⋅ ⋅ ⋅= =⋅ ⋅ ⋅ ⋅

PC = 56 hands

49. There are 13 clubs in a deck. (13, 5) 13 12 11 10 9

(13, 5)5! 5 4 3 2 1

⋅ ⋅ ⋅ ⋅= =⋅ ⋅ ⋅ ⋅

PC

= 1287 hands



5-17

50. There are 26 red cards in a deck. (26, 5)

(26, 5)5!

26 25 24 23 22

5 4 3 2 165,780 hands

=

⋅ ⋅ ⋅ ⋅=⋅ ⋅ ⋅ ⋅

=

PC

51. (20, 3) 20 19 18

(20, 3)3! 3 2 1

⋅ ⋅= =⋅ ⋅

PC

= 1140 ways

52. Let n be the number of appetizers. ( , 5)

( , 5)5!

( –1)( – 2)( – 3)( – 4)

120

=

=

P nC n

n n n n n

> 700 n(n – 1)(n – 2)(n – 3)(n – 4) > 84,000

This is close to 5( – 2) 84,000,>n so try n close

to 52 84,000 12+ ≈ (rounded to the nearest

whole number). (12, 5)

(12, 5)5!

12 11 10 9 8

5 4 3 2 1792700

=

⋅ ⋅ ⋅ ⋅=⋅ ⋅ ⋅ ⋅

=>

PC

Check when n = 11: (11, 5)

(11, 5)5!

11 10 9 8 7

5 4 3 2 1462700

=

⋅ ⋅ ⋅ ⋅=⋅ ⋅ ⋅ ⋅

=<

PC

Thus the least possible number of appetizers is 12.

53. P(5, 5) = 5 · 4 · 3 · 2 · 1 = 120 ways

54. P(6, 3) = 6 · 5 · 4 = 120 signals

55. (8, 4) 8 7 6 5

(8, 4) 704! 4 3 2 1

⋅ ⋅ ⋅= = =⋅ ⋅ ⋅

PC ways

56. Starting with the digits 9876543210 which are strictly decreasing from left to right, remove any four digits. The remaining six digits will be in decreasing order. The number of ways this can be done is C(10,4) = 210. For example, if you decide to remove 8,7,3, 2, the number reads 965410, which is decreasing.

57. Moe has (9, 2) 9 8

(9, 2) 362! 2 1

⋅= = =⋅

PC choices.

Joe has (7, 3) 7 6 5

(7, 3) 353! 3 2 1

⋅ ⋅= = =⋅ ⋅

PC choices.

Thus Joe is correct.

58. Consider the problem as each basketball franchise, in order, selecting a city. Then there are P(5, 3) = 5 · 4 · 3 = 60 ways.

59. This is the same as ordering 12 different pictures. Then there are P(12, 12) = 12! = 479,001,600 arrangements.

60. P(9, 2) = 9 · 8 = 72 exacta bets

C(9, 2) = ( )9, 2 72

2! 2 1=

⋅P

= 36 quinella bets

61. There are P(5, 5) = 5 · 4 · 3 · 2 · 1 = 120 ways to order the mathematics books. Counting the group of mathematics books as one item, there are P(5, 5) = 120 ways to order the novels and the set of mathematics books. All together, there are 120 · 120 = 14,400 ways to arrange these books.

62. 12 · 2 · 1 · 9 · 2 · 1 · 6 · 2 · 1 · 3 · 2 · 1 = 31,104 ways

63. P(7, 4) = 7 · 6 · 5 · 4 = 840 arrangements

64. Among the four players who can play at any of the positions catcher, first base, third base, and shortstop, there are P(4, 4) = 4! = 24 possibilities. Among the four players who can play any of the three outfield positions or second base, there are also P(4, 4) = 4! = 24 possibilities. There are 5 possibilities for pitcher. Thus there are 24 · 24 · 5 = 2880 ways the manager can assign players to positions.

65. There is only one possibility for the pitcher and there are two possibilities for the first baseman. Of the remaining seven players, there are P(7, 7) = 7! = 5040 possibilities. Thus there are 1 · 2 · 5040 = 10,080 batting orders.



5-18

66. There are (14, 2) 14 13

(14, 2) 912! 2 1

⋅= = =⋅

PC

ways to select two comedians. There are

(20, 3) 20 19 18(20, 3) 1140

3! 3 2 1

⋅ ⋅= = =⋅ ⋅

PC ways to

select three singers. Thus he can select the acts in 91 · 1140 = 103,740 ways.

67. P(5, 5) = 5! = 120 ways

68. (7, 3) 7 6 5

(7, 3) 353! 3 2 1

⋅ ⋅= = =⋅ ⋅

PC ways

69. Of cards from two suits, there are (26, 5) 26 25 24 23 22

(26, 5)5! 5 4 3 2 1

⋅ ⋅ ⋅ ⋅= =⋅ ⋅ ⋅ ⋅

PC

= 65,780 five-card combinations. Of those,

(13, 5) 13 12 11 10 9(13, 5)

5! 5 4 3 2 1

⋅ ⋅ ⋅ ⋅= =⋅ ⋅ ⋅ ⋅

PC

= 1287 consist of cards from one of the two suits.

Since there are (4, 2) 4 3

(4, 2) 62! 2 1

⋅= = =⋅

PC

possible two-suit combinations, there are 6 ⋅ (65,780 − 2 ⋅ 1287) = 379,236 five-card combinations from exactly two suits.

70. Using the hint, there are 7 spots for the A’s and 4 of them must be chosen. This can be done in

( ) ( )7, 4 7 6 5 47, 4

4! 4 3 2 1

⋅ ⋅ ⋅= =⋅ ⋅ ⋅

PC = 35 ways.

71. The product is even if at least one of the digits is

even. There are 510 100,000= possible ZIP

codes altogether. There are 55 3125= possible ZIP codes consisting of all odd digits. Thus there are 100,000 – 3125 = 96,875 ZIP codes in which the product of the digits is even.

72. (6, 2)

3 (6,2) 3 3 15 45games2!

⋅ = ⋅ = ⋅ =PC

73. If there are n people at the party, the number of

possible handshakes is ( 1)

.2

−n n So if there were

45 handshakes, then ( 1)

452

− =n n or

n(n − 1) = 90. The solution is n = 10.

74. Let n = number of colors. number of single color jerseys + number of two-color jerseys ≥ 25 n + C(n, 2) ≥ 25

n n + C(n, 2)

2 3

3 6

4 10

5 15

6 21

7 28

The correct answer is (c).

75. There are C(15, 3) = 455 different side dishes in which all sides are different. There are 2 ⋅ C(15, 2) = 2(105) = 210 different ways to choose two same side dishes and one different side. (For example if you choose the two sides A and B, there are C(15,2) ways to choose which sides you want. There are two ways to choose which side will be duplicated- for instance AAB or ABB. Finally, there are C(15,1) = 15 ways to choose 3 servings of one side dish. The total is 455 + 210 + 15 = 680.

76. The number of ways to choose three scoops of ice cream with all different flavors is C(16,3) = 560. The number of ways to choose three scoops of ice cream with two flavors the same is 2(C(16,2) = 240. The number of ways to choose three scoops of ice cream with all three scoops the same is C(16,1) = 16. Total number of ways is 560 + 240 + 16 = 816.

77. The number of possible combinations of six numbers chosen from 1 to 49 is C(49, 6) = 13,983,816; if the numbers are chosen from 1 to 59 this increases to C(59, 6) = 45,057,474. The ratio is

(59, 6)3.22,

(49, 6)

C

C so the correct answer is (b).

78. At 220 tickets per week, it would take

approximately (59, 6)

204,807 weeks220

C or

204,8073939 years,

52 so the correct answer is

(d).



5-19

79. A cube has 8 vertices, so one can construct C(8, 3) = 56 distinct triangles.

80. ( , ) ( 1)( 2) ( 1)( 1)( 2) ( 1)( )( 1) 3 2 1

( )( 1) 3 2 1!

( )!

= − − − +− − − + − − − ⋅ ⋅=

− − − ⋅ ⋅

=−

P n r n n n n rn n n n r n r n r

n r n rn

n r

81. Number of programs for a student on the semester system: C(752, 5) · C(747, 5) · C(742, 5) · C(737, 5) ⋅ C(732, 5) · C(727, 5) · C(722, 5) · C(717, 5) Number of programs for a student on the trimester system: C(937, 3) · C(934, 3) · C(931, 3) · C(928, 3) · C(925, 3) · C(922, 3) · C(919, 3) · C(916, 3) · C(913, 3) · C(910, 3) · C(907, 3) · C(904, 3) Using a computer to compare the two quantities, semester system students have a greater number of different programs.

82. a. P(50, 50) = 50! ways

b. There are P(10, 10) = 10! ways to arrange the first column and there are P(40, 40) = 40! ways to arrange the rest. Thus there are 10!40! arrangements.

c. There are P(40, 10) ways to arrange the first column. From the remaining chips, there are P(40, 40) = 40! ways to arrange the rest. Thus, there are P(40, 10) · 40! arrangements.

83. a. C(45, 5) = 1,221,759 tickets

b. C(100, 4) = 3,921,225 tickets

c. First lottery

d. Number of possible tickets for the first lottery: P(45, 5) = 146,611,080 Number of possible tickets for the second lottery: P(100, 4) = 94,109,400 Thus the second lottery is better if order matters.

Exercises 5.6

1. a. 62 64= outcomes

b. 6 5 4

(6, 3) 203 2 1

⋅ ⋅= =⋅ ⋅

C outcomes

c. These are the outcomes with exactly no tails, one tail, or two tails. 6 6 5

(6, 0) (6, 1) (6, 2) 1 221 2 1

⋅+ + = + + =⋅

C C C outcomes

d. These are all of the outcomes minus the outcomes with exactly no heads or one head.

6 62 – (6, 0) – (6, 1) 64 –1 57

1= − =C C outcomes



5-20

2. Two of the seven blocks in the route must be “south.”

7 6(7, 2) 21

2 1

⋅= =⋅

C routes

3. Four of the nine blocks in the route must be “south.”

9 8 7 6(9, 4) 126

4 3 2 1

⋅ ⋅ ⋅= =⋅ ⋅ ⋅

C routes

4. a. 12 11 10 9

(12, 4) 4954 3 2 1

⋅ ⋅ ⋅= =⋅ ⋅ ⋅

C samples

b. 8 7 6 5

(8, 4) 704 3 2 1

⋅ ⋅ ⋅= =⋅ ⋅ ⋅

C samples

c. 8 7 4 3

(8, 2) (4, 2)2 1 2 1

⋅ ⋅= ⋅⋅ ⋅

C C

= 28 · 6 = 168 samples

d. Samples with exactly three red balls: 8 7 6 4

(8, 3) (4, 1) 56 43 2 1 1

⋅ ⋅= ⋅ = ⋅⋅ ⋅

C C

= 224 samples From part (b) there are 70 samples with all red balls. Samples with at least 3 red balls: 224 + 70 = 294

e. Samples with exactly the same number of red and white balls:

8 7 4 3(8, 2) (4, 2)

2 1 2 128 6168 samples

⋅ ⋅= ⋅⋅ ⋅

= ⋅=

C C

Samples that contain a different number of red balls than white balls: 495 – 168 = 327 samples

5. a. 10 9 8

(10, 3) 1203 2 1

⋅ ⋅= =⋅ ⋅

C samples

b. 8 7 6

(8, 3) 563 2 1

⋅ ⋅= =⋅ ⋅

C samples

c. From part (b) there are 56 samples with no rotten apples. Samples that contain at least one rotten apple: 120 – 56 = 64 samples

6. a. 82 256= outcomes

b. 8 7 6

(8, 3) 563 2 1

⋅ ⋅= =⋅ ⋅

C outcomes

c. Outcomes with exactly no heads or one head: 8

(8, 0) (8, 1) 1 91

+ = + =C C outcomes

Outcomes with at least two heads: 256 – 9 = 247 outcomes

d. 8 7 6 5 8 7 6 5 4

(8, 4) (8, 5)4 3 2 1 5 4 3 2 1

⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅+ = +⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅

C C

= 70 + 56 = 126 outcomes

7. Consider this a sequence of three operations. Assign 25 students to dorm A: C(100, 25) Assign 40 of the remaining 75 students to dorm B: C(75, 40) Assign 35 of the remaining 35 students to dorm C: C(35, 35) = 1 Ways to assign all 100 students:

44(100, 25) (75, 40) 7.14 10⋅ ≈ ×C C

8. There are two possible winners. Observe that for a series to end in n games the winning team must win the nth game. Thus the winning team must win 3 of the first n – 1 games. Ways to win in four games: C(3, 3) = 1 way Ways to win in five games:

4 3 2(4, 3) 4

3 2 1

⋅ ⋅= =⋅ ⋅

C ways

Ways to win in six games: 5 4 3

(5, 3) 103 2 1

⋅ ⋅= =⋅ ⋅

C ways

Ways to win in seven games: 6 5 4

(6, 3) 20 ways3 2 1

⋅ ⋅= =⋅ ⋅

C

Total number of ways to win a series: 1 + 4 + 10 + 20 = 35 ways Number of sequences: 2 · 35 = 70 sequences

9. Two of the five blocks from A to C must be “south.” Routes from A to C:

5 4(5, 2) 10

2 1

⋅= =⋅

C routes

Two of the four blocks from C to B must be “south.”



5-21

Routes from C to B: 4 3

(4, 2) 62 1

⋅= =⋅

C routes

Routes from A to B passing through C: 10 · 6 = 60 routes

10. a. 100 99 98 97 96

(100, 5)5 4 3 2 1

⋅ ⋅ ⋅ ⋅=⋅ ⋅ ⋅ ⋅

C

= 75,287,520 samples

b.

90 89 87 10 9(90, 3) (10, 2)

3 2 1 2 1117,480 455,286,600 samples

⋅ ⋅ ⋅⋅ = ⋅⋅ ⋅ ⋅

= ⋅=

C C

c. Samples with no defective fuses: 90 89 88 87 86

(90, 5)5 4 3 2 1

⋅ ⋅ ⋅ ⋅=⋅ ⋅ ⋅ ⋅

C

= 43,949,268 samples Samples with at least 1 defective fuse: 75,287,520 – 43,949,268 = 31,338,252 samples

11. First select the states from which each senator comes. There are

50 49 48 47 46(50, 5)

5 4 3 2 12,118,760 ways to do this.

⋅ ⋅ ⋅ ⋅=⋅ ⋅ ⋅ ⋅

=

C

Once the states have been selected, each state has two possible choices of senators, so there

are 52 32= possible committees once the states have been chosen. Ways a committee of five senators can be selected so that no two members are from the same state: 2,118,760 · 32 = 67,800,320 ways

12. Ways with exactly 3 correct answers: 5 4 3

(5, 3) 103 2 1

⋅ ⋅= =⋅ ⋅

C ways

Ways with exactly 4 correct answers: 5 4 3 2

(5, 4) 54 3 2 1

⋅ ⋅ ⋅= =⋅ ⋅ ⋅

C ways

Ways with exactly 5 correct answers: 5 4 3 2 1

(5, 5) 15 4 3 2 1

⋅ ⋅ ⋅ ⋅= =⋅ ⋅ ⋅ ⋅

C way

Ways with 3 or more correct answers: 10 + 5 + 1 = 16 ways (Note that this is exactly half of the 32 ways since it should equal the number of ways answering 3 or more incorrectly, which is the complement of the set.)

13. 4 3 2 6 5 4

(4, 3) (6, 3)3 2 1 3 2 1

⋅ ⋅ ⋅ ⋅⋅ = ⋅⋅ ⋅ ⋅ ⋅

C C

4 2080 ways

= ⋅=

14. First arrange the three categories on the shelf. There are 3! = 6 ways to do this. Then arrange the two novels. There are 2! = 2 ways to do this. Then arrange the six biographies. There are 6! = 720 ways to do this. Then arrange the four how-to books. There are 4! = 24 ways to do this. Ways to arrange the books: 6 · 2 · 720 · 24 = 207,360 ways

15. C(20, 12) · C(8, 2) = 125,970 · 28 = 3,527,160 ways

16. C(4, 2) · C(5, 2) = 6 · 10 = 60 ways

17. C(9, 5) · C(8, 6) = 126 · 28 = 3528 ways

18. C(10, 5) · C(5, 4) · C(1, 1) = 252 · 5 · 1 = 1260 ways

19. C(15, 3) · C(12, 8) · C(4, 4) = 455 · 495 · 1 = 225,225 ways

20. C(10, 7) · C(3, 3) = 120 · 1 = 120 ways

21. P(26, 6) = 26 · 25 · 24 · 23 · 22 · 21 = 165,765,600 plates

22. C(15, 5) · C(10, 4) = 3003 · 210 = 630,630 ways

23. P(7, 3) = 7 · 6 · 5 = 210 words

24. First select the two consonants in the sequence.

There are 21 20

(21, 2) 2102 1

⋅= =⋅

C ways to do this.

Then arrange the seven letters. There are 7! = 5040 arrangements. Number of sequences: 210 · 5040 = 1,058,400 sequences

25. Step 1) There are 10 letters in the word “ABSTEMIOUS.” Step 2) Choose 5 slots to the put the letters “AEIOU” in that order. There are C(10,5) ways to do this. Step 3) There are P(5,5) number of ways to arrange the five remaining letters. Total: C(10, 5) ⋅ P(5, 5) = 252(120) = 30,240.



5-22

26. C(20, 5) · C(15, 4) · C(11, 3) = 15,504 · 1365 · 165 = 3,491,888,400 ways

27. First select the lineup of the four classes. There are 4! = 24 ways to do this. Then arrange three of the four freshmen. There are P(4, 3) = 24 ways to do this. Then arrange three of the five sophomores. There are P(5, 3) = 60 ways to do this. Then arrange three of the six juniors. There are P(6, 3) = 120 ways to do this. Then arrange three of the seven seniors. There are P(7, 3) = 210 ways to do this. Number of different pictures: 24 · 24 · 60 · 120 · 210 = 870,912,000 pictures

28. 5! = 120

29. First select the order of the towns. There are 2! = 2 ways to do this. Then select the order in the larger town. There are 3! = 6 ways to do this. Then select the order in the smaller town. There are 2! = 2 ways to do this. Ways to schedule his stops: 2 · 6 · 2 = 24 ways

30. C(14, 3) · C(11, 4) · C(7, 5) = 364 · 330 · 21 = 2,522,520 ways

31. C(4, 3) · C(4, 2) = 4 · 6 = 24 hands

32. First select the two aces. There are C(4, 2) = 6 ways to do this. Then select the second denomination. There are 12 ways to do this. Then select two of the denomination. There are C(4, 2) = 6 ways to do this. Then select the third denomination. There are 11 ways to do this. Then select one of the denomination. There are 4 ways to do this. Number of poker hands: 6 · 12 · 6 · 11 · 4 = 19,008 hands

33. First select the first denomination. There are 13 ways to do this. Then select three of the denomination. There are C(4, 3) = 4 ways to do this. Then select the second denomination. There are 12 ways to do this. Then select two of the denomination. There are C(4, 2) = 6 ways to do this. Number of poker hands: 13 · 4 · 12 · 6 = 3744 hands

34. First select the denominations of the two pairs. There are C(13, 2) = 78 ways to do this. Then select two of one denomination. There are C(4, 2) = 6 ways to do this. Then select two of the other denomination. There are C(4, 2) = 6 ways to do this. Then select the third denomination. There are 11 ways to do this. Then select one of the denomination. There are 4 ways to do this. Number of poker hands: 78 · 6 · 6 · 11 · 4 = 123,552 hands

35. The digits that can be used in a detour-prone ZIP code are 0, 1, 6, 8 and 9. There are

55 3125= ZIP codes that use only these digits. However, some of these are the same when read upside down and so are not detour-prone. For this to happen, the middle digit must be 0, 1 or 8 and the last two digits read upside down must match the first two, as in for example 98186. There are 5 ⋅ 5 ⋅ 3 = 75 ways this can happen, so 3125 − 75 = 3050 ZIP codes are detour prone.

36. Consider the case with no restrictions. First assign 2 students to the first double. There are C(6, 2) = 15 ways to do this. Then assign 2 students to the second double. There are C(4, 2) = 6 ways to do this. Then assign students to the singles. There are 2! = 2 ways to do this. No restrictions: 15 · 6 · 2 = 180 ways Next consider the case where the students who cannot be placed together are in the first double. There are C(4, 2) · C(2, 1) · C(1, 1) = 6 · 2 · 1 = 12 ways to do this. Likewise, if they are in the second double, there are also 12 ways to do this. Assign with restriction: 180 – 12 – 12 = 156 ways

37. Committees with no restrictions: C(12, 4) = 495 committees Committees with no seniors: C(9, 4) = 126 committees Committees with at least one senior: 495 – 126 = 369 committees

38. C(15, 5) = 3003 ways

39. Consider the case with 3 tap, 1 ballet, and 1 modern routine. There are C(8, 3) · C(5, 1) · C(2, 1) = 56 · 5 · 2 = 560 ways to do this. Consider the case with 1 tap, 3 ballet, and 1 modern routine. There are C(8, 1) · C(5, 3) · C(2, 1) = 8 · 10 · 2 = 160 ways to do this.



5-23

Consider the case with 2 tap, 2 ballet, and 1 modern routine. There are C(8, 2) · C(5, 2) · C(2, 1) = 28 · 10 · 2 = 560 ways to do this. Consider the case with 2 tap, 1 ballet, and 2 modern routines. There are C(8, 2) · C(5, 1) · C(2, 2) = 28 · 5 · 1 = 140 ways to do this. Consider the case with 1 tap, 2 ballet, and 2 modern routines. There are C(8, 1) · C(5, 2) · C(2, 2) = 8 · 10 · 1 = 80 ways to do this. Thus the manager can choose 560 + 160 + 560 + 140 + 80 = 1500 ways.

40. Choose whether a man or woman sits at the leftmost end. There are 2 choices. Then order the men. There are 6! = 720 ways to do this. Then order the women. There are 6! = 720 ways to do this. Thus there are a total of 2 · 720 · 720 = 1,036,800 ways to seat them.

41. First order the men. There are 6! = 720 ways to do this. Then order the women. There are 6! = 720 ways to do this. Since no two women can sit next to each other, there must be at least one man between each two. This is a total of five men. The sixth man can be at the end or with one of the other men. There are 7 ways to do this. Thus there are a total of 720 · 720 · 7 = 3,628,800 ways to seat them.

42. There are C(6, 3) = 20 ways for exactly three of them to come to dinner. There are C(6, 3) + C(6, 4) + C(6, 5) + C(6, 6) = 20 + 15 + 6 + 1 = 42 ways for three or more of them to come to dinner.

43. Each family member can decide to come or

not come to dinner. Thus there are 62 64= different groups that can come to dinner. (This includes no one coming to dinner.)

44. 6! = 720 ways

45. 3 blue marbles and no white marbles chosen: C(3, 3) · C(5, 0) = 1 · 1 = 1 way 2 blue marbles and 1 white marble chosen: C(3, 2) · C(5, 1) = 3 · 5 = 15 ways number of blue marbles chosen exceeds the number of white marbles chosen: 1 + 15 = 16 ways

46. 10 5

252

⋅ = (Divide by 2 since a handshake consists

of two hands.)

47. There are 426 456,976= possible four-letter

sequences; of these, 421 194,481= contain no

vowels. Hence 456,976 − 194,481 = 262,495 are possible words.

48. a. C(38, 22) = 22,239,974,430 classes

b. C(20, 12) · C(18, 10) = 5,512,195,260 classes

49. Total number of possible bridge hands: C(52, 13) Bridge hands with all four aces: C(48, 9) Percentage of bridge hands with all four aces:

(48, 9).00264 .264%

(52, 13)≈ =C

C

50. Bridge hands with all four aces: C(48, 9) = 1,677,106,640 Bridge hands with exactly two kings and two queens: C(4, 2) · C(4, 2) · C(44, 9) = 25,521,498,288 A bridge hand with exactly two kings and two queens is more likely.

51. Bridge hands with all four aces: C(48, 9) = 1,677,106,640 Bridge hands with the two red kings, the two red queens, and no other kings or queens: C(44, 9) = 708,930,508 A bridge hand with all four aces is more likely.

Exercises 5.7

1. 6 56

(6, 2) 152 2 1

⋅⎛ ⎞ = = =⎜ ⎟ ⋅⎝ ⎠C

2. 7 6 57

(7, 3) 353 3 2 1

⋅ ⋅⎛ ⎞ = = =⎜ ⎟ ⋅ ⋅⎝ ⎠C

3. 88

(8, 1) 81 1⎛ ⎞ = = =⎜ ⎟⎝ ⎠

C

4. 9

(9, 9) 19⎛ ⎞ = =⎜ ⎟⎝ ⎠

C

5. 18 1718

(18, 16) (18, 2) 15316 2 1

⋅⎛ ⎞ = = = =⎜ ⎟ ⋅⎝ ⎠C C



5-24

6. 2525

(25, 24) (25, 1) 2524 1⎛ ⎞ = = = =⎜ ⎟⎝ ⎠

C C

7. 7

(7, 0) 10⎛ ⎞ = =⎜ ⎟⎝ ⎠

C

8. 66

(6, 1) 61 1⎛ ⎞ = = =⎜ ⎟⎝ ⎠

C

9. 8

(8, 8) 18⎛ ⎞ = =⎜ ⎟⎝ ⎠

C

10. 9

(9, 0) 10⎛ ⎞ = =⎜ ⎟⎝ ⎠

C

11. ( , –1) ( , 1)–1 1

⎛ ⎞ = = = =⎜ ⎟⎝ ⎠

nnC n n C n n

n

12. ( , ) 1⎛ ⎞ = =⎜ ⎟⎝ ⎠

nC n n

n

13. 0! = 1

14. 1! = 1

15. n · (n – 1)! = n!

16. ! ( –1)!

( –1)!⋅= =n n n

nn n

17.

6

6 6 6 6 6 6 60 1 2 3 4 5 6

264

⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞+ + + + + +⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠==

18.

7

7 7 7 7 7 7 7 70 1 2 3 4 5 6 7

2128

⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞+ + + + + + +⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠==

19. 10 9 8 2

10 9 8 2

10 10 100 1 2

10 45

⎛ ⎞ ⎛ ⎞ ⎛ ⎞+ +⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠= + +

x x y x y

x x y x y

20. 20 19 18 2

20 19 18 2

20 20 200 1 2

20 190

⎛ ⎞ ⎛ ⎞ ⎛ ⎞+ +⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠= + +

x x y x y

x x y x y

21. 2 13 14 15

2 13 14 15

15 15 1513 14 15

105 15

⎛ ⎞ ⎛ ⎞ ⎛ ⎞+ +⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠= + +

x y xy y

x y xy y

22. 2 10 11 12

2 10 11 12

12 12 1210 11 12

66 12

⎛ ⎞ ⎛ ⎞ ⎛ ⎞+ +⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠= + +

x y xy y

x y xy y

23. 10 10 10 1020184,756

10⎛ ⎞ =⎜ ⎟⎝ ⎠

x y x y

24. 5 5 5 510252

5⎛ ⎞ =⎜ ⎟⎝ ⎠

x y x y

25. 11

3307

⎛ ⎞ =⎜ ⎟⎝ ⎠

26. 13

12875

⎛ ⎞ =⎜ ⎟⎝ ⎠

27. 62 64= subsets

28. 1002 subsets

29. 42 16= tips

30. 42 16= types

31. 52 32= options (The number of subsets of any size taken from a set of five elements.)

32. For dressings, there are six choices, including no dressing.

There are 62 6 384⋅ = possible salads.

33. 82 1 255− = ways

34. 62 1 63− = ways

35. 152 3 2 196,608⋅ ⋅ = types

36. 72 (7, 0) (7, 1) 128 1 7120 ways

− − = − −=

C C

37. 72 (7, 6) (7, 7) 128 7 1120 ways

− − = − −=

C C



5-25

38. There are 4 choices of ice cream and 52 1 31 ways− = to choose a selection of toppings (since at least one topping

must be selected), and hence 4 ⋅ 31 = 124 possible sundaes. An alternate interpretation is that at most one sauce and at most one type of nut is chosen; in this case, there are 4 choices of sauce (including no sauce) and 3 choices of nuts (including no nuts). This gives 4 ⋅ 3 − 1 = 11 choices of toppings (since at least one topping must be selected) and hence 4 ⋅ 11 = 44 possible sundaes.

39. Find the number of subsets of the set {a, b, d, e}. 42 16= subsets

40. Find the number of subsets of the set {1, 3, 5}. 32 8= subsets

41. 82 – (8, 0) – (8, 1) 256 –1– 8247 ways

==

C C

42. The number of subsets with an odd number of elements is C(n, 1) + C(n, 3) + C(n, 5) + … + C(n, n – 2) + C(n, n), which is equal to C(n, n – 1) + C(n, n – 3) + C(n, n – 5) + … + C(n, 2) + C(n, 0), the number of subsets with an even number of elements.

43. 122 4096= groups

44. (7, 2) (10, 3) (5, 1) 21 120 512,600 menus

⋅ ⋅ = ⋅ ⋅=

C C C

45. Each region contains either elements in no sets, exactly one set, exactly two sets, or all three sets. The number of

regions containing elements in no set is 3

.0⎛ ⎞⎜ ⎟⎝ ⎠

The number of regions containing elements in exactly one set is

3.

1⎛ ⎞⎜ ⎟⎝ ⎠

The number of regions containing elements in exactly two sets is 3

.2⎛ ⎞⎜ ⎟⎝ ⎠

The number of regions containing

elements in all three sets is 3

.3⎛ ⎞⎜ ⎟⎝ ⎠

Thus, there are 3 3 3 3

1 3 3 1 8 regions.0 1 2 3⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞+ + + = + + + =⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠

46. (20, 1) (20, 2) (20, 3) (20, 4) (20, 5)C C C C C+ + + +

= 20 + 190 + 1140 + 4845 + 15,504 = 21,699 ways

47. C(8, 3) · C(12, 2) = 56 · 66 = 3696 ways

48. Not including the choice of no sales: 202 –1 1,048,575= ways

49. 7 7( – 3 ) [ (–3 )]= +x y x y

The term with 4y is 3 4 3 4 3 47(–3 ) 35 81 2835 .

4⎛ ⎞ = ⋅ =⎜ ⎟⎝ ⎠

x y x y x y The coefficient of 4y is 32835 .x



5-26

50. a. 5 5 5 5 5 50 2 4 5 3 1⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞+ + = + +⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠

by equation (4), so 5 5 5 5 5 50 1 2 3 4 50.

⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞− + − + −⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠=

b. The same reasoning applies when n is odd:

0 2 4 1 3 5⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞+ + + = + + +⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠

n n n n n n

so 0.0 1 2 3⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞− + − + − =⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠

n n n n nn

c. 50 [1 ( 1)]5 5 5 5 5 50 1 2 3 4 5

= + −⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞= − + − + −⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠

d. For any n:

0 [1 ( 1)]

( 1)0 1 2 3

= + −⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞= − + − + + −⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠

n

nn n n n nn

51. Find the number of subsets of size 1 or more from a set of 15 elements. The total number of subsets of a set of 15 elements (including the

null set) is 152 . Thus, if the empty set is not included, the total number of sets of one or more

elements is 152 1 32,767.− =

Exercises 5.8

1. 5!

203!1!1!

=

2. 5!

302!1!2!

=

3. 6!

1802!1!2!1!

=

4. 6!

203!3!

=

5. 7!

2103!2!2!

=

6. 7!

1054!1!2!

=

7. 12!

34,6504!4!4!

=

8. 8!

5603!3!2!

=

9. 12!

166,3205!3!2!2!

=

10. 8!

25202!2!2!2!

=

11. 5

1 15!1,401,400

5! (3!)⋅ =

12. 2

1 10!126

2! (5!)⋅ =

13. 3

1 18!2,858,856

3! (6!)⋅ =

14. 3

1 12!5775

3! (4!)⋅ =

15. 20!20

99,768,2407, 5, 8 7!5!8!⎛ ⎞ = =⎜ ⎟⎝ ⎠

reports

16. 15!15

756,756 ways5, 5, 5 5!5!5!⎛ ⎞ = =⎜ ⎟⎝ ⎠

17. 5

1 20!2,546,168,625

5! (4!)⋅ = ways

18. 30!30

10, 2, 18 10!2!18!⎛ ⎞ =⎜ ⎟⎝ ⎠

= 5,708,552,850 ways

19. 4

1 20!488,864,376

4! (5!)⋅ = ways

20. 9!9

16803, 3, 3 3!3!3!⎛ ⎞ = =⎜ ⎟⎝ ⎠

ways

21. 4!4

12 ways1, 1, 2 1!1!2!⎛ ⎞ = =⎜ ⎟⎝ ⎠



5-27

22. 10!10

31504, 4, 2 4!4!2!⎛ ⎞ = =⎜ ⎟⎝ ⎠

ways

23. 7

1 14!135,135

7! (2!)⋅ = ways

24. First partition the six days traveling and four

days at the office. There are 10

2106, 4⎛ ⎞ =⎜ ⎟⎝ ⎠

ways

to do this without any restriction. Of those, 7 have 4 consecutive days at the office. So she has 203 ways to partition her travel and nontravel days. Next partition the six travel days among the three cities. There are

6!690

2, 2, 2 2!2!2!⎛ ⎞ = =⎜ ⎟⎝ ⎠

ways to do this. Thus

there are 203 · 90 = 18,270 ways for her to schedule her travel.

25. The number of ways ten students are to be divided into two five member teams for a

basketball game is 2

10!126.

(2)!(5!)=

26. Select the number of elements of 1 :S

11 1 1

!

!( – )!⎛ ⎞= =⎜ ⎟⎝ ⎠

nnp

n n n n ways

Select the number of elements of 2 :S

112

2 2 1 2

( – )!–

!( – – )!⎛ ⎞= =⎜ ⎟⎝ ⎠

n nn np

n n n n n ways

Select the number of elements of (1 ) :< <kS k m

1 2 –1– – – –⎛ ⎞= ⎜ ⎟⎝ ⎠

kk

k

n n n np

n

1 2 –1

1 2 –1

( – – – – )!

!( – – – – – )!= k

k k k

n n n n

n n n n n n ways

Select the number of elements of :mS

1 2 –1– – –

! way

!

−⎛ ⎞= ⎜ ⎟⎝ ⎠⎛ ⎞= ⎜ ⎟⎝ ⎠

=

mm

m

m

m

m

m

n n n np

n

nn

n

n

Observe that for 1 < k < m, the numerator of Pk

is 1 1( )!−− − − kn n n which also occurs in the

denominator of 1−kp , which is

1 1 1!( )!− −− − −k kn n n n . Using the generalized

multiplication principle, the number of ordered partitions is 1 2⋅ ⋅ ⋅ mp p p . All of the

numerators cancel in this multiplication for 1 < k ≤ m, leaving 1 2! ! !⋅ ⋅ ⋅ mn n n in the

denominator. Thus,

1 21 2 3

!.

! ! ! !⋅ ⋅ ⋅ =

⋅ ⋅ ⋅ ⋅mm

np p p

n n n n

27. 38

10, 12, 10, 638!

10!12!10!6!

⎛ ⎞⎜ ⎟⎝ ⎠

=

= 115,166,175,166,136,334,240 ways

28. 415

65!8.81 10

(13!)= × ways

Chapter 5 Supplementary Exercises

1. { } { } { }, , , ,∅ a b a b

2. ( )′ ′ ′∪ = ∩S T S T

S T

U

3. ( ) 16!16, 2 120 possibilities

2!14!= =C

4. 2 · 5! = 240 ways

5.

R S

T

U

6.

12 11 10 2

12 11 10 2

12 12 120 1 2

12 66

⎛ ⎞ ⎛ ⎞ ⎛ ⎞+ +⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠= + +

x x y x y

x x y x y

7. ( ) ( ) 8! 6!8, 3 6, 2 56 15 840

3!5! 2!4!⋅ = ⋅ = ⋅ =C C

8. Let U = {people}, P = {people who received placebos}, I = {people who showed improvement}. n(U) = 60; n(P) = 15; n(I) = 40; ( )′ ∩n P I = 30

Draw and complete a Venn diagram as shown.



5-28

IP30

105

15

U

( )′ ′∩n P I = 15

Fifteen of the people who received the drug showed no improvement.

9. 7 · 5 = 35 combinations

10. 12!12

13,8602, 4, 6 2!4!6!⎛ ⎞ = =⎜ ⎟⎝ ⎠

11. Let U = {applicants}, F = {applicants who speak French}, S = {applicants who speak Spanish}, and G = {applicants who speak German}. n(U) = 115; n (F) = 70; n(S) = 65; n(G) = 65; ( ) ( ) ( )45; 35; 40;∩ = ∩ = ∩ =n F S n S G n F G

n(F ∩ S ∩ G) = 35. Draw and complete a Venn diagram.

G

F

5

20S

025

1035

20

0

U

(( ) ) 0′∪ ∪ =n F S G

None of the people speak none of the three languages.

12. 17 1617 17

13615 2 2 1

⋅⎛ ⎞ ⎛ ⎞= = =⎜ ⎟ ⎜ ⎟ ⋅⎝ ⎠ ⎝ ⎠

For Exercises 13–20, let U = {members of the Earth Club}, W = {members who thought the priority is clean water}, A = {members who thought the priority is clean air}, R = {members who thought the priority is recycling}. Then n(U) = 100, n(W) = 45, n(A) = 30, n(R) = 42, n(W ∩ A) = 13, n(A ∩ R) = 20, n(W ∩ R) = 16, n(W ∩ A ∩ R) = 9. Draw and complete the Venn diagram as follows.

R

W

7

25A

1115

49

6

23

U

13. ( ) 6′ ′∩ ∩ =n A W R

14. ( ) ( )( ) ( ) ( )25 7 6 11 32 17 49∩ ′ ∪ ′ ∩ = + + + = + =n W A W A

15. ( )( ) 25 15 7 47∪ ∩ ′ = + + =n W R A

16. ( ) 11∩ ∩ ′ =n A R W

17. ( ) ( ) ( )( ) 25 6 15 46∩ ′ ∩ ′ ∪ ′ ∩ ∩ ′ ∪ ′ ∩ ′ ∩ = + + =n W A R W A R W A R

18. ( ) 23 25 6 4 58′ = + + + =n R



5-29

19. ( ) 15 7 22∩ ′ = + =n R A

20. ( )( ) 23∪ ∪ ′ =n W A R

21. C(9, 4) = C(9, 5) = 126

22. 402

23. Let S = {students who ski}, H = {students who play ice hockey}. Then

( ) ( ) ( ) ( )400 300 150550.

n S H n S n H n S H∪ = + − ∩= + −=

24. 6 · 10 · 8 = 480 meals

25. 5!5

20 ways1, 3, 1 1!3!1!⎛ ⎞ = =⎜ ⎟⎝ ⎠

26. The first digit can be anything but 0, the hundreds digit must be 3, and the last digit must be even.

49 1 5 10 450,000⋅ ⋅ ⋅ =

27. 2 89 10 8,100,000,000⋅ =

28. 2 3 1026 26 26 26 146,813,779,479,510+ + + + =

29. Strings of length 8 formed from the symbols a, b, c, d, e:

85 390,625= strings Strings of length 8 formed from the symbols a, b, c, d:

84 65,536= strings Strings with at least one e: 390,625 – 65,536 = 325,089 strings

30. ( ) 12!12, 5 792

5!7!= =C

31. ( ) 100!100, 14

14!86!44,186,942,677,323,600 groups

=

=

C

32. U = {households} F = {households that get Fancy Diet Magazine} C = {households that get Clean Living Journal}

( ) ( ) ( ) ( )4000 10,000 150012,500

∪ = + − ∩= + −=

n F C n F n C n F C

( )( ) ( ) ( )40,000 12,50027,500

∪ ′ = − ∪= −=

n F C n U n F C

27,500 households get neither.

33. 103 59,049= paths

34. ( ) 60!60, 10 75,394,027,566

10!50!= =C ways

35. 105 9,765,625= tests

36. 621 85,766,121= strings

37. ( ) 10!10, 4 210

4!6!= =C ways

38. 3

1 21!66,512,160 ways

3! (7!)⋅ =

39. 14! = 87,178,291,200 ways

40. ( )5

1 100!

5! 20!⋅ ways

41. Suppose one of the senators from New York has been assigned to a group. There are C(98, 19) ways to complete this group.

( ) 98! 20 80 98!98, 19

19!79! 20!80!

⋅ ⋅= =C

To assign the remaining 80 senators into groups of

20 each, there are ( )4

1 80!

4! 20!⋅ ways.

Thus the number of ways to divide the senators if the 2 senators from the state of New York cannot be in the same group is as follows.



5-30

( )( ) ( )

( )

( )

( )

( )

( )

( )

4 4

5

5

5

5

5

5

1 80! 20 80 98! 1 80!98, 19

4! 20!80! 4!20! 20!

20 80 98!

4! 20!5 20 80 98!

5 4! 20!100 80 98!

5! 20!80 100 99 98!

99 5! 20!80 100!

99 5! 20!80 1 100!

99 5! 20!

⎛ ⎞⋅ ⋅⎛ ⎞⎜ ⎟⋅ ⋅ = ⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠⋅ ⋅=

⋅ ⋅ ⋅=⋅

⋅ ⋅=

⋅ ⋅ ⋅=⋅⋅=

⋅

= ⋅ ⋅

C

Thus there are ( )5

80 1 100!

99 5! 20!⋅ ⋅ ways to divide the

senators. 80

This is of the answer in Exercise 40.99

⎛ ⎞⎜ ⎟⎝ ⎠

42. 3 ⋅ C(10, 2) = 135

43. A diagonal corresponds to a pair of vertices, except that adjacent pairs must be excluded. Hence there are

( 1) ( 3)( , 2)

2 2

− −− = − =n n n nC n n n diagonals.

44. 8 ⋅ 6 = 48

45. 5! · 4! · 3! · 2! · 1! = 120 · 24 · 6 · 2 · 1 = 34,560

46. P(12, 5) = 12 · 11 · 10 · 9 · 8 = 95,040

47. There are four suits. Once a suit is chosen, select 5 of the 13 cards. Poker hands with cards of the same suit: 4 · C(13, 5) = 4 · 1287 = 5148 hands

48. 9 · 9 · 8 = 648 numbers

49. Exactly the first two digits alike: 9 · 1 · 9 = 81 First and last digit alike: 9 · 9 · 1 = 81 Last two digits alike: 9 · 9 · 1 = 81 Numbers with exactly two digits alike: 81 + 81 + 81 = 243

50. C(4, 3) · C(48, 2) = 4 · 1128 = 4512 hands

51. 24 23 24 23 22 552 12,14412,696 names

⋅ + ⋅ ⋅ = +=

52. 3! = 6 pairings

53. 2 · 3! · 3! = 72 ways

54. 3 · 2 · 2 · 1 = 12 ways

55. Any two lines will intersect, so the number of intersections is C(10, 2) = 45. Since each of the ten lines is a boundary there are ten sides to the feasible set and there are ten vertices. The number of intersections that occur outside the feasible set is 45 – 10 = 35.

56. First teacher: ( )4

1 24!96,197,645,544

4! 6!⋅ = ways

Second teacher:

( )6

1 24!4,509,264,634,875

6! 4!⋅ = ways

The second teacher has more options.

57. 10!10

42003, 4, 3 3!4!3!⎛ ⎞ = =⎜ ⎟⎝ ⎠

58. Let n be the number of books. n! = 120, so n = 5. There are five books.

59. 7 · 6 · 5 · 1 · 4 · 3 · 2 · 1 · 1 = 5040 orders

60. a. C(12, 2) · C(12, 3) = 66 · 220 = 14,520

b. Select five out of twelve couples. Then determine the spouse from each couple.

( ) 512, 5 2 25,344⋅ =C

61. C(7, 2) + C(7, 1) + C(7, 0) = 29 ways

62. There is one possible path from A at the top to the first B on the left, and 1 possible path from there to the final C. Similarly, for the last B on the right. For the second and fourth B’s, there are 4 incoming and 4 outgoing paths for the third B, there are 6 incoming and 6 outgoing paths. so the total number of paths from A at the top of C or

the bottom is 2 2 2 2 21 4 6 4 1 70.+ + + + =

63. There are 22 26⋅ 3-letter call letters and 32 26⋅

4-letter call letters, so 2 32 26 2 26 36,504⋅ + ⋅ = call letters are possible.



5-31

64. Find n such that n! = 479,001,600. Testing numbers on computer, n = 12.

65. 25

10, 9, 6⎛ ⎞⎜ ⎟⎝ ⎠

=16,360,143,800 ways

66. a. First choose the three letters. There are C(26, 3) ways to do this. Then choose the three numbers. There are C(10, 3) ways to do this. Then order the six. There are 6! ways to do this. The number of license plates is C(26, 3) · C(10, 3) · 6! = 224,640,000.

b. The number of license plates consisting of three distinct letters followed by three distinct numbers is

(26, 3) (10, 3) (26, 3) 3! (10, 3) 3!= ⋅ ⋅ ⋅P P C C .

Thus there are ( ) ( )

( ) ( )26, 3 10, 3 6!

2026, 3 3! 10, 3 3!

⋅ ⋅=

⋅ ⋅C C

C C times as

many plates in (a).

67. If A ∩ B = ∅ then A and B have no elements in common.

Conceptual Exercises

68. If n(A ∪ B) = n(A) + n(B), then A and B are disjoint sets and have no elements in common.

69. n(n − 1)! = n(n − 1)(n − 2)(n − 3)...1 = n! For example, 5 ⋅ 4! = 5(4 ⋅ 3 ⋅ 2 ⋅ 1) = 5 ⋅ 4 ⋅ 3 ⋅ 2 ⋅ 1 = 5!.

70. Let n = 1. We have 1(1-1)! = 1(0)! = 1! = 1 So, since 1 times any number is the number, we have 0! = 1! = 1.

71. A permutation is an arrangement of items in which order is important. A combination is a subset of objects taken from a larger set in which the order of the objects does not make a difference.

72. The number of committees of size 6 is the same as the number of committees of size 4. For each committee of size 6, there is a committee consisting of the four people who were not chosen. For example, use letters A, B,C, D, E, F, G, H, I, J to represent the people. For the committee {A, B, C, D, E, F} the complementary committee is {G, H, I, J}.

73. C(10, 3) = C(10, 7). The number of subsets of size 3 taken from a 10 member set is the same as the number of subsets of size 7. For example, consider the set N = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}. For the subset A = {1, 2, 3}, there is the corresponding subset B = {4, 5, 6, 7, 8, 9, 10}.

74. A committee of size 5 either includes or excludes John Doe. If the committee includes him, there are C(10, 4) ways to choose the other 4 members. If the committee excludes him, there are C(10, 5) ways to choose the 5 members. Hence C(10, 4) + C(10, 5) = C(11, 5).

Chapter 5 Chapter Test

1. a. 4! = 4 · 3 · 2 · 1 = 24

b. P(7, 3) = 7 · 6 · 5 = 210

c. ( ) 18 1718, 16 153

2 1

⋅= =⋅

C

d. 6!6

10 0!6!⎛ ⎞ = =⎜ ⎟⎝ ⎠

e. 5!5

302, 1, 2 2!1!2!⎛ ⎞ = =⎜ ⎟⎝ ⎠

2. a. True

b. True

c. False

3. a. { , }′ ∩ =S T a e

b. ( )∪ ′ = ∅S T

4. ∩C E represents the set of certified public accountants who are self-employed.

′∪C E Represents the set of people who are either certified public accountants or are not self-employed.

5.

BA

C



5-32

6. Let C = {members of Choral Society} and D = {members of Drama Club}. Then ∪C D = {member of Choral Society or Drama Club} and ∩C D = {members of Choral Society and Drama Club}.

n(C) = 40, n (D) = 30, ( )∪n C D = 60

( )∪n C D = n(C) + n(D) – ( )∩n C D

60 = 40 + 30 – ( )∩n C D

( )∩n C D = 40 + 30 – 60 = 10

10 students are members of both groups.

7. Let E = {education majors}, M = {middle-of-the-road politically}, and V = {those who did volunteer work}. n(E) = 9, n(M) = 43, n(V) = 82 n(E ∩ M) = 4, n(E ∩ V) = 6, n(M ∩ V) = 32, n(E ∩ M ∩ V) = 3

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )9 43 82 4 6 32 395

n E M V n E n M n V n E M n E V n M V n E M V∪ ∪ = + + − ∩ − ∩ − ∩ + ∩ ∩= + + − − − +=

So 100 − 95 = 5 freshman did not meet any of the three criteria.

8. 3 · 5 · 4 = 60 ways

9. a. ( ) ( ) 20! 30!20, 5 30, 5

15!5! 25!5!⋅ = ⋅C C

= (15,504)(142,506) = 2,209,413,024 ways

b. ( ) 10!10, 6 210

6!4!= =C ways

Note that this is the same as C(10, 4).

10. 7 · 6 · 5 = 210 numbers

11. Consider the children as a set, so there are 3 items to arrange. This can be done in 3! = 6 ways. For each of those arrangements, the children can be arranged in 4! = 24 ways. The total number of arrangements is 6 · 24 = 144.

12. ( ) ( ) 10! 20!10, 3 20, 3

7!3! 17!3!120 1140136,800 ways

⋅ = ⋅

= ⋅=

C C

13. 12!12

7925 7!5!

⎛ ⎞ = =⎜ ⎟⎝ ⎠

14. 72 128=

15. 8!8

25202, 2, 2, 2 2!2!2!2!⎛ ⎞ = =⎜ ⎟⎝ ⎠

ways

Chapter 5 Project

1. n = 5, r = 2, and r = 3



5-33

2. ( )

( ) ( ) ( )1 ! ( 1)!1 1

1 ![( 1) ]1 ! 1 1 !

− −− −⎛ ⎞ ⎛ ⎞+ = +⎜ ⎟ ⎜ ⎟− − −− − − −⎡ ⎤⎝ ⎠ ⎝ ⎠ ⎣ ⎦

n nn nr r r n rr n r

( )( ) ( )

( )( )

( )( )

( )( )( )

( ) ( )( )( )

( ) [ ]( )

( )( )

1 ! 1 !

1 ! ! ! 1 !

1 ! 1 !

! ! ! !

1 ! 1 !

! !

1 !

! !!

! !

,

− −= +

− − − −− − −

= +− −

− + − −=

−− + −

=−

=−

=⎛ ⎞= ⎜ ⎟⎝ ⎠

n n

r n r r n r

r n n r n

r n r r n r

r n n r n

r n r

n r n r

r n rn

r n r

C n r

nr

3. Using the hint, there are C(n – 1, r – 1) ways to select r objects including x because this is equivalent to selecting r – 1 from the remaining n – 1 objects. The number of ways to select a subset of r objects not containing x is C(n – 1, r) because r objects must be selected from the n – 1 objects that are not x. The total number of ways to do

this is ( ) ( ) 1 11, 1 1,

1− −⎛ ⎞ ⎛ ⎞− − + − = +⎜ ⎟ ⎜ ⎟−⎝ ⎠ ⎝ ⎠

n nC n r C n r

r r.

4. th

th

th

th

th

th

1 7 21 35 35 21 7 1 7 row

1 8 28 56 70 56 28 8 1 8 row

1 9 36 84 126 126 84 36 9 1 9 row

1 10 45 120 210 252 210 120 45 10 1 10 row

1 11 55 165 330 462 462 330 165 55 11 1 11 row

1 12 66 220 495 792 924 792 495 220 66 12 1 12 row

12 12792, 924

5 6⎛ ⎞ ⎛ ⎞= =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

5. a.

( ) 1 2 2 12 1 1 1 1 1 1 1 1 1 10 1 2 1

0 1 2 3

− − −⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞= + = + ⋅ + ⋅ +…+ ⋅ +⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟−⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞= + + + +…+⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠

nn n n n n nn n n n nn n

n n n n nn

b.

( ) ( ) ( ) ( )2 11 20 0 1 1 1 1 1 1 1 1 1 ( 1)0 1 2 1

0 1 2 3

−− −⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞= = + − = + − + − +…+ ⋅ − + −⎡ ⎤ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎣ ⎦ −⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞= − + − +…±⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠

n nn n n n nn n n n nn n

n n n n nn

c. 7 7 7 7

1 21 35 7 640 2 4 6⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞+ + + = + + + =⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠

7 7 7 77 35 21 1 64

1 3 5 7⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞+ + + = + + + =⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠



5-34

d. Part (b) shows that the sum of the even-numbered elements equals the sum of the odd-numbered elements. If n is even, then

00 1 2 3 1

so .0 2 1 3 1

⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞− + − +…− + =⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟−⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞+ +…+ = + +…+⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟−⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠

n n n n n nn n

n n n n n nn n

If n is odd, then

00 1 2 3 1

so .0 2 1 1 3

⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞− + − +…+ − =⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟−⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞+ +…+ = + +…+⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟−⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠

n n n n n nn n

n n n n n nn n

The common sum is 1

(2 )2

n , or 12 −n .

e. Since there are an equal number of even and odd-numbered subsets, a set S with n elements has 2n subsets so

that half of them, or 122 ,

2−=

nn have an even number of elements.

6. a. Let +1

n

1

1 2 4 + 8 + + 2

2 2 4 + 8 + 16 + + 2

( 1 + 2 + 4 + 8 + + 2 )

1 2 +

= + += +

− == − +

n

n

n

S

S

S

S

Therefore 11 2 4 8 2 2 1++ + + +…+ = −n n .

b. From 5(a) we know that the sum of the nth row is 2n . Rearranging the equation in 6(a), we get

( )( )st

1

The sum ofThe sum of all previous rowsthe 1 row

1 2 4 8 2 1 2 +↑

+

+ + + +…+ + =n n

n

7. a. n = 2, 3, 5, 7, 11

b. ( )( ) ( )1 2 1

1 2 3

− − … − +⎛ ⎞ =⎜ ⎟ ⋅ ⋅ ⋅…⋅⎝ ⎠

p p p p rpr r

Since the result is always an integer, p(p – 1)(p – 2) · … · (p – r + 1) must be divisible by 1 2 3 !.⋅ ⋅ ⋅ ⋅ =r r But p is prime and p > r, so

(p – 1)(p – 2) · … · (p – r + 1) must be divisible by r!. Call the result of this division rm . Then ,⎛ ⎞ =⎜ ⎟⎝ ⎠ r

pm p

r

and this is divisible by p.

c. The sum of all of the numbers in the pth row is 2p (by 5(a)). There are 2 numbers in the row that are not

interior numbers, and they are both 1. Therefore the sum of the interior number is 2 2−p .

d. 72 2 126− = 126

187

=



5-35

e. By 7(c), for any prime number p, the sum of the interior numbers of the pth row is 2 2−p . By 7(b), each of these numbers is divisible by p, so their sum must also be divisible by p. Thus, for any prime number p,

2 2−p is a multiple of p.

8.

0

1

1

2

1

2

2

3

1

2

2

3

2

number of row

odd numbers

0 1 2

1 2 2

2 2 2

3 4 2

4 2 2

5 4 2

6 4 2

7 8 2

8 2 2

9 4 2

10 4 2

11 8 2

12 4 2

=============

9. Answers may vary. Sample answer: Each row of the small triangle has a number of dots that is equal to a power of 2. These small triangles then appear in multiples of 2 in the larger triangle. Since the original power of 2 is always multiplied by a power of 2, the result is always a power of 2.


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