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Section 10.6 Permutations and Combinations 10-1 10.6 Permutations and Combinations In this section, we will learn to 1. Use the Multiplication Principle for Events. 2. Solve permutation problems. 3. Solve combination problems. 4. Use combination notation to write the Binomial Theorem. 5. Find the number of permutations of like things. 1. Use the Multiplication Principle for Events Lydia plans to go to dinner and attend a movie. If she has a choice of four restau- rants and three movies, in how many ways can she spend her evening? There are four choices of restaurants and, for any one of these choices, there are three choices of movies, as shown in the tree diagram in Figure 10-3. Lydia’s choices Cheesecake Factory Outback Steakhouse P.F. Chang’s Olive Garden True Grit Chronicles of Narnia: The Voyage of the Dawn Treader Black Swan True Grit Chronicles of Narnia: The Voyage of the Dawn Treader Black Swan True Grit Chronicles of Narnia: The Voyage of the Dawn Treader Black Swan True Grit Chronicles of Narnia: The Voyage of the Dawn Treader Black Swan FIGURE 10-3 Not For Sale © Cengage Learning. All rights reserved. No distribution allowed without express authorization.
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Page 1: © Cengage Learning. All rights reserved. No distribution ... · Section 10.6 Permutations and Combinations 10-1 10.6 Permutations and Combinations In this section, we will learn

Section10.6PermutationsandCombinations10-1

10.6 Permutations and CombinationsInthissection,wewilllearnto

1.UsetheMultiplicationPrincipleforEvents.2.Solvepermutationproblems.3.Solvecombinationproblems.4.UsecombinationnotationtowritetheBinomialTheorem.5.Findthenumberofpermutationsoflikethings.

1. Use the Multiplication Principle for eventsLydia plans to go to dinner and attend a movie. If she has a choice of four restau-rants and three movies, in how many ways can she spend her evening? There are four choices of restaurants and, for any one of these choices, there are three choices of movies, as shown in the tree diagram in Figure 10-3.

Lydia’schoices

Cheesecake Factory

Outback Steakhouse

P.F. Chang’s

Olive Garden

True GritChronicles of Narnia: The Voyage of the Dawn TreaderBlack SwanTrue GritChronicles of Narnia: The Voyage of the Dawn TreaderBlack SwanTrue GritChronicles of Narnia: The Voyage of the Dawn TreaderBlack SwanTrue GritChronicles of Narnia: The Voyage of the Dawn TreaderBlack Swan

FigUre 10-3Not For Sale

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Page 2: © Cengage Learning. All rights reserved. No distribution ... · Section 10.6 Permutations and Combinations 10-1 10.6 Permutations and Combinations In this section, we will learn

10-2 Chapter10Sequences,Series,andProbability

Multiplication Principle for events

The diagram shows that Lydia has 12 ways to spend her evening. One possibil-ity is to eat at the Cheesecake Factory and watch True Grit. Another is to eat at the Olive Garden and watch Black Swan.

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Any situation that has several outcomes is called an event. Lydia’s first event (choosing a restaurant) can occur in 4 ways. Her second event (choosing a movie) can occur in 3 ways. Thus, she has 4 ? 3, or 12, ways to spend her evening. This example illustrates the Multiplication Principle for Events.

Let E1 and E2 be two events. If E1 can be done in a1 ways, and if—after E1 has occurred—E2 can be done in a2 ways, then the event “E1 followed by E2” can be done in a1 ? a2 ways.

The Multiplication Principle can be extended to n events.

eXaMPLe 1 Using the Multiplication Principle for events

If a traveler has 4 ways to go from New York to Chicago, 3 ways to go from Chicago to Denver, and 6 ways to go from Denver to San Francisco, in how many ways can she go from New York to San Francisco?

sOLUtiOn We can let E1 be the event “going from New York to Chicago,” E2 the event “going from Chicago to Denver,” and E3 the event “going from Denver to San Francisco.” Since there are 4 ways to accomplish E1, 3 ways to accomplish E2, and 6 ways to accomplish E3, the number of routes available is

4 ? 3 ? 6 5 72

self Check 1 If a man has 4 sweaters and 5 pairs of slacks, how many different outfits can he wear?

now try exercise 29.

2. solve Permutation ProblemsSuppose we want to arrange 7 books in order on a shelf. We can fill the first space with any one of the 7 books, the second space with any of the remaining 6 books,

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Page 3: © Cengage Learning. All rights reserved. No distribution ... · Section 10.6 Permutations and Combinations 10-1 10.6 Permutations and Combinations In this section, we will learn

Section10.6PermutationsandCombinations10-3

Formula for P 1n, r2

the third space with any of the remaining 5 books, and so on, until there is only one space left to fill with the last book. According to the Multiplication Principle, the number of ordered arrangements of the books is

7 ? 6 ? 5 ? 4 ? 3 ? 2 ? 1 5 5,040

When finding the number of possible ordered arrangements of books on a shelf, we are finding the number of permutations. The number of permutations of 7 books, using all the books, is 5,040. The symbol P 1n, r2 is read as “the number of permuta-tions of n things r at a time.” Thus, P 17, 72 5 5,040.

eXaMPLe 2 Finding the number of Permutations

Assume that there are 7 signal flags of 7 different colors to hang on a mast. How many different signals can be sent when 3 flags are used?

sOLUtiOn We are asked to find P 17, 32 , the number of permutations (ordered arrangements) of 7 things using 3 of them. Any one of the 7 flags can hang in the top position on the mast. Any one of the 6 remaining flags can hang in the middle position, and any one of the remaining 5 flags can hang in the bottom position. By the Multiplication Principle, we have

P 17, 32 5 7 ? 6 ? 5 5 210

It is possible to send 210 different signals.

self Check 2 How many signals can be sent if two of the 7 flags are missing?

now try exercise 37.

Although it is correct to write P 17, 32 5 7 ? 6 ? 5, we will change the form of the answer to obtain a convenient formula. To derive this formula, we proceed as follows:

P 17, 32 5 7 ? 6 ? 5

57 ? 6 ? 5 ? 4 ? 3 ? 2 ? 1

4 ? 3 ? 2 ? 1 Multiply numerator and denominator by 4 ? 3 ? 2 ? 1.

57!

4!

57!

17 2 32 !The generalization of this idea gives the following formula.

The number of permutations of n things r at a time is given by

P 1n, r2 5n!

1n 2 r2 !

eXaMPLe 3 Using the Permutation Formula

Find: a. P 18, 42 b. P 1n, n2 c. P 1n, 02 sOLUtiOn We will substitute into the formula for P 1n, r2 and simplify.

a. P 18, 42 58!

18 2 42 ! 58 ? 7 ? 6 ? 5 ? 4!

4!5 1,680Not For Sale

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Page 4: © Cengage Learning. All rights reserved. No distribution ... · Section 10.6 Permutations and Combinations 10-1 10.6 Permutations and Combinations In this section, we will learn

10-4 Chapter10Sequences,Series,andProbability

Formulas for P 1n, n2 and P 1n, 02

b. P 1n, n2 5n!

1n 2 n2 ! 5n!

0!5

n!

15 n!

c. P 1n, 02 5n!

1n 2 02 ! 5n!

n!5 1

self Check 3 Find: a. P 17, 52 b. P 16, 02 now try exercise 13.

Parts (b) and (c) of Example 3 establish the following formulas.

The number of permutations of n things n at a time and n things 0 at a time are given by the formulas

P 1n, n2 5 n! and P 1n, 02 5 1

eXaMPLe 4 solving a Permutation Problem

In how many ways can a baseball manager arrange a batting order of 9 players if there are 25 players on the team?

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sOLUtiOn To find the number of permutations of 25 things 9 at a time, we substitute 25 for n and 9 for r in the formula for finding P 1n, r2 .

P 1n, r2 5n!

1n 2 r2 !

P 125, 92 525!

125 2 92 !

525!

16!

525 ? 24 ? 23 ? 22 ? 21 ? 20 ? 19 ? 18 ? 17 ? 16!

16!

5 741,354,768,000

The number of permutations is 741,354,768,000.

self Check 4 In how many ways can the manager arrange a batting order if 2 players can’t play?

now try exercise 41.

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Page 5: © Cengage Learning. All rights reserved. No distribution ... · Section 10.6 Permutations and Combinations 10-1 10.6 Permutations and Combinations In this section, we will learn

Section10.6PermutationsandCombinations10-5

aCCent On teChnOLOgy PermutationsA graphing calculator can be used to find P 1n, r2 . For example, let’s consider the permutation in Example 4 and determine P 125, 92 . The calculator operation that will be used is nPr.

to find P 125, 92, first enter 25 in your calculator.

Press MATH and move the cursor to PrB.scroll down to 2:nPr and press ENTER .

enter 9 on your calculator and press ENTER .We see that P 125, 92 is equal to 741,354,768,000.

FigUre 10-4

eXaMPLe 5 solving a Permutation Problem

In how many ways can 5 people stand in a line if 2 people refuse to stand next to each other?

sOLUtiOn The total number of ways that 5 people can stand in line is

P 15, 52 5 5! 5 5 ? 4 ? 3 ? 2 ? 1 5 120

To find the number of ways that 5 people can stand in line if 2 people insist on standing together, we consider the two people as one person. Then there are 4 people to stand in line, and this can be done in P 14, 42 5 4! 5 24 ways. However, because either of the two who are paired could be first, there are two arrangements for the pair who insist on standing together. Thus, there are 2 ? 4!, or 48 ways that 5 people can stand in line if 2 people insist on standing together.

The number of ways that 5 people can stand in line if 2 people refuse to stand together is 5! 5 120 (the total number of ways to line up 5 people) minus 2 ? 4! 5 48 (the number of ways to line up the 5 people if 2 do stand together):

120 2 48 5 72

There are 72 ways to line up the people.

self Check 5 In how many ways can 5 people stand in a line if one person demands to be first?

now try exercise 39.

eXaMPLe 6 solving a Permutation Problem

In how many ways can 5 people be seated at a round table?

sOLUtiOn If we were to seat 5 people in a row, there would be 5! possible arrangements. How-ever, at a round table, each person has a neighbor to the left and to the right. If each person moves one, two, three, four, or five places to the left, everyone has the same neighbors and the arrangement has not changed. Thus, we must divide 5! by 5 to get rid of these duplications. The number of ways that 5 people can be seated at a round table is

5!

55

5 ? 4!

55 4! 5 4 ? 3 ? 2 ? 1 5 24Not For Sale

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Page 6: © Cengage Learning. All rights reserved. No distribution ... · Section 10.6 Permutations and Combinations 10-1 10.6 Permutations and Combinations In this section, we will learn

10-6 Chapter10Sequences,Series,andProbability

Formulas for C 1n, n2 and C 1n, 02

Formula for C 1n, r2

Circular arrangements

self Check 6 In how many ways can 6 people be seated at a round table?

now try exercise 43.

The results of Example 6 suggest the following fact.

There are 1n 2 12 ! ways to arrange n things in a circle.

3. solve Combination ProblemsSuppose that a class of 12 students selects a committee of 3 persons to plan a party. With committees, order is not important. A committee of John, Maria, and Raul is the same as a committee of Maria, Raul, and John. However, if we assume for the moment that order is important, we can find the number of permutations of 12 things 3 at a time.

P 112, 32 512!

112 2 32 ! 512 ? 11 ? 10 ? 9!

9!5 1,320

However, since we do not care about order, this result of 1,320 ways is too large. Because there are 6 ways 13! 5 62 of ordering every committee of 3 students, the result of P 112, 32 5 1,320 is exactly 6 times too big. To get the correct number of committees, we must divide P 112, 32 by 6:

P 112, 326

51,320

65 220

In cases of selection where order is not important, we are interested in combi-nations, not permutations. The symbols C 1n, r2 and an

rb both mean the number of

combinations of n things r at a time.If a committee of r people is chosen from a total of n people, the number of

possible committees is C 1n, r2 , and there will be r! arrangements of each committee. If we consider the committee as an ordered grouping, the number of orderings is P 1n, r2 . Thus, we have

(1) r!C 1n, r2 5 P 1n, r2We can divide both sides of Equation 1 by r! to obtain the formula for finding C 1n, r2 .

C 1n, r2 5 anrb 5

P 1n, r2r!

5n!

r! 1n 2 r2 !

The number of combinations of n things r at a time is given by

C 1n, r2 5 anrb 5

n!

r! 1n 2 r2 !

In Exercises 77 and 78, you will be asked to prove the following formulas.

If n is a whole number, then

C 1n, n2 5 1 and C 1n, 02 5 1

Comment

Whendiscussingpermutations,ordercounts.Whendiscussingcombinations,orderdoesn’tcount.

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Page 7: © Cengage Learning. All rights reserved. No distribution ... · Section 10.6 Permutations and Combinations 10-1 10.6 Permutations and Combinations In this section, we will learn

Section10.6PermutationsandCombinations10-7

aCCent On teChnOLOgy

eXaMPLe 7 solving a Combination Problem

If Carla must read 4 books from a reading list of 10 books, how many choices does she have?

sOLUtiOn Because the order in which the books are read is not important, we find the number of combinations of 10 things 4 at a time:

C 110, 42 510!

4! 110 2 42 ! 510 ? 9 ? 8 ? 7 ? 6!

4 ? 3 ? 2 ? 1 ? 6!

510 ? 9 ? 8 ? 7

4 ? 3 ? 2

5 210

Carla has 210 choices.

self Check 7 How many choices would Carla have if she had to read 5 books?

now try exercise 49.

CombinationsA graphing calculator can be used to find C 1n, r2 . For example, let’s consider the permutation in Example 7 and determine C 110, 42 . The calculator operation that will be used is nCr.

to find C 110, 42, first enter 10 in your calculator.

Press MATH and move the cursor to PrB.scroll down to 3:nCr and press ENTER .

enter 4 on your calculator and press ENTER .We see that C 110, 42 is equal to 210.

FigUre 10-5

eXaMPLe 8 solving a Combination Problem

A class consists of 15 men and 8 women. In how many ways can a debate team be chosen with 3 men and 3 women?

sOLUtiOn There are C 115, 32 ways of choosing 3 men and C 18, 32 ways of choosing 3 women. By the Multiplication Principle, there are C 115, 32 ? C 18, 32 ways of choosing mem-bers of the debate team:

C 115, 32 ? C 18, 32 515!

3! 115 2 32 ! ?8!

3! 18 2 32 !

515 ? 14 ? 13

6?

8 ? 7 ? 66

5 25,480

There are 25,480 ways to choose the debate team.Not For Sale

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Page 8: © Cengage Learning. All rights reserved. No distribution ... · Section 10.6 Permutations and Combinations 10-1 10.6 Permutations and Combinations In this section, we will learn

10-8 Chapter10Sequences,Series,andProbability

Binomial theorem

self Check 8 In how many ways can the debate team be chosen if it is to have 4 men and 2 women?

now try exercise 59.

4. Use Combination notation to Write the Binomial theoremThe formula

C 1n, r2 5n!

r! 1n 2 r2 !gives the coefficient of the 1r 1 12 th term of the binomial expansion of 1a 1 b2 n. This implies that the coefficients of a binomial expansion can be used to solve prob-lems involving combinations. The Binomial Theorem is restated below—this time listing the 1r 1 12 th term and using combination notation.

If n is any positive integer, then

1a 1 b2 n 5 an0ban 1 an

1ban21b 1 an

2ban22b2 1c1 an

rban2rbr 1c1 an

nbbn

Intheexpansionof 1a 1 b2 n,thetermcontainingbrisgivenby

anrban2rbr

eXaMPLe 9 Using Pascal’s triangle to Compute a Combination

Use Pascal’s Triangle to compute C 17, 52 . sOLUtiOn Consider the eighth row of Pascal’s Triangle and the corresponding combinations:

1 7 21 35 35 21 7 1

a70b a7

1b a7

2b a7

3b a7

4b a7

5b a7

6b a7

7b

C 17, 52 5 a75b 5 21.

self Check 9 Use Pascal’s Triangle to compute C 16, 52 . now try exercise 75.

5. Find the number of Permutations of Like thingsA word is a distinguishable arrangement of letters. For example, six words can be formed with the letters a, b, and c if each letter is used exactly once. The six words are

abc, acb, bac, bca, cab, and cba

If there are n distinct letters and each letter is used once, the number of distinct words that can be formed is n! 5 P 1n, n2 . It is more complicated to compute the number of distinguishable words that can be formed with n letters when some of the letters are duplicates.

Comment

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Page 9: © Cengage Learning. All rights reserved. No distribution ... · Section 10.6 Permutations and Combinations 10-1 10.6 Permutations and Combinations In this section, we will learn

Section10.6PermutationsandCombinations10-9

Permutations of Like things

eXaMPLe 10 Finding the number of Permutations of Like things

Find the number of “words” that can be formed if each of the 6 letters of the word little is used once.

sOLUtiOn For the moment, we assume that the letters of the word little are distinguish-able: “LitTle.” The number of words that can be formed using each letter once is 6! 5 P 16, 62 . However, in reality we cannot tell the l’s or the t’s apart. Therefore, we must divide by a number to get rid of these duplications. Because there are 2! order-ings of the two l’s and 2! orderings of the two t’s, we divide by 2! ? 2!. The number of words that can be formed using each letter of the word little is

P 16, 622! ? 2!

56!

2! ? 2!5

6 ? 5 ? 42

? 3 ? 2 ? 12 ? 1 ? 2 ? 1

5 180

self Check 10 How many words can be formed if each letter of the word balloon is used once?

now try exercise 55.

Example 10 illustrates the following general principle.

The number of permutations of n things with a things alike, b things alike, and so on, is

n!

a!b!c

self Check answers 1. 20 2. 60 3. a. 2,520 b. 1 4. 296,541,907,200 5. 24 6. 120 7. 252 8. 38,220 9. 6 10. 1,260

Exercises 10.6getting readyYou should be able to complete these vocabulary and concept statements before you proceed to the practice exercises.

Fill in the blanks. 1. If E1 and E2 are two events and E1 can be done in

4 ways and E2 can be done in 6 ways, then the event E1 followed by E2 can be done in ways.

2. An arrangement of n objects is called a .

3. P 1n, r2 5

4. P 1n, n2 5

5. P 1n, 02 5

6. There are ways to arrange n things in a circle.

7. C 1n, r2 5

8. Using combination notation, C 1n, r2 5 .

9. C 1n, n2 5 10. C 1n, 02 5

11. If a word with n letters has a of one letter, b of another letter, and so on, the number of different

words that can be formed is .

12. Where the order of selection is not important, we are interested in , not .

PracticeEvaluate each expression. 13. P 17, 42 14. P 18, 32 15. C 17, 42 16. C 18, 32 17. P 15, 52 18. P 15, 02 19. a5

4b 20. a8

4b

21. a50b 22. a5

5b

23. P 15, 42 ? C 15, 32 24. P 13, 22 ? C 14, 32 Not For Sale

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Page 10: © Cengage Learning. All rights reserved. No distribution ... · Section 10.6 Permutations and Combinations 10-1 10.6 Permutations and Combinations In this section, we will learn

10-10 Chapter10Sequences,Series,andProbability

25. a53b a4

3b a3

3b 26. a5

5b a6

6b a7

7b a8

8b

27. a6866

b 28. a10099

b

applications 29. Choosing lunch A lunchroom has a machine with

eight kinds of sandwiches, a machine with four kinds of soda, a machine with both white and chocolate milk, and a machine with three kinds of ice cream. How many different lunches can be chosen? (Consider a lunch to be one sandwich, one drink, and one ice cream.)

30. Manufacturing license plates How many six-digit license plates can be manufactured if no license plate number begins with 0?

31. Available phone numbers How many different seven-digit phone numbers can be used in one area code if no phone number begins with 0 or 1?

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32. Arranging letters In how many ways can the letters of the word number be arranged?

33. Arranging letters with restrictions In how many ways can the letters of the word number be arranged if the e and r must remain next to each other?

34. Arranging letters with restrictions In how many ways can the letters of the word number be arranged if the e and r cannot be side by side?

35. Arranging letters with repetitions How many ways can five Scrabble tiles bearing the letters, F, F, F, L, and U be arranged to spell the word fluff ?

36. Arranging letters with repetitions How many ways can six Scrabble tiles bearing the letters B, E, E, E, F, and L be arranged to spell the word feeble?

37. Placing people in line In how many arrangements can 8 women be placed in a line?

38. Placing people in line In how many arrangements can 5 women and 5 men be placed in a line if the women and men alternate?

39. Placing people in line In how many arrangements can 5 women and 5 men be placed in a line if all the men line up first?

40. Placing people in line In how many arrangements can 5 women and 5 men be placed in a line if all the women line up first?

41. Combination locks How many permutations does a combination lock have if each combination has 3 numbers, no two numbers of the combination are the same, and the lock dial has 30 notches?

42. Combination locks How many permutations does a combination lock have if each combination has 3 numbers, no two numbers of the combination are the same, and the lock dial has 100 notches?

43. Seating at a table In how many ways can 8 people be seated at a round table?

44. Seating at a table In how many ways can 7 people be seated at a round table?

45. Seating at a table In how many ways can 6 people be seated at a round table if 2 of the people insist on sitting together?

46. Seating arrangements with conditions In how many ways can 6 people be seated at a round table if 2 of the people refuse to sit together?

47. Arrangements in a circle In how many ways can 7 children be arranged in a circle if Sally and John want to sit together and Martha and Peter want to sit together?

48. Arrangements in a circle In how many ways can 8 children be arranged in a circle if Laura, Scott, and Paula want to sit together?

49. Selecting candy bars In how many ways can 4 candy bars be selected from 10 different candy bars?

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50. Selecting birthday cards In how many ways can 6 birthday cards be selected from 24 different cards?

51. Circuit wiring A wiring harness containing a red, a green, a white, and a black wire must be attached to a control panel. In how many different orders can the wires be attached?

52. Grading homework A professor grades home-work by randomly checking 7 of the 20 problems assigned. In how many different ways can this be done?

53. Forming words with distinct letters How many words can be formed from the letters of the word plastic if each letter is to be used once?

54. Forming words with distinct letters How many words can be formed from the letters of the word computer if each letter is to be used once?

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Section10.6PermutationsandCombinations10-11

55. Forming words with repeated letters How many words can be formed from the letters of the word banana if each letter is to be used once?

56. Forming words with repeated letters How many words can be formed from the letters of the word laptop if each letter is to be used once?

57. Manufacturing license plates How many license plates can be made using two different letters fol-lowed by four different digits if the first digit cannot be 0 and the letter O is not used?

58. Planning class schedules If there are 7 class periods in a school day, and a typical student takes 5 classes, how many different time patterns are possible for the student?

59. Selecting golf balls From a bucket containing 6 red and 8 white golf balls, in how many ways can we draw 6 golf balls of which 3 are red and 3 are white?

60. Selecting a committee In how many ways can you select a committee of 3 Republicans and 3 Demo-crats from a group containing 18 Democrats and 11 Republicans?

61. Selecting a committee In how many ways can you select a committee of 4 Democrats and 3 Republi-cans from a group containing 12 Democrats and 10 Republicans?

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62. Drawing cards In how many ways can you select a group of 5 red cards and 2 black cards from a deck containing 10 red cards and 8 black cards?

63. Planning dinner In how many ways can a husband and wife choose 2 different dinners from a menu of 17 dinners?

64. Placing people in line In how many ways can 7 people stand in a row if 2 of the people refuse to stand together?

65. Geometry How many lines are determined by 8 points if no 3 points lie on a straight line?

66. Geometry How many lines are determined by 10 points if no 3 points lie on a straight line?

67. Coaching basketball How many different teams can a basketball coach start if the entire squad consists of 10 players? (Assume that a starting team has 5 players and each player can play all positions.)

68. Managing baseball How many different teams can a manager start if the entire squad consists of 25 players? (Assume that a starting team has 9 players and each player can play all positions.)

69. Selecting job applicants There are 30 qualified applicants for 5 openings in the sales department. In how many different ways can the group of 5 be selected?

70. Sales promotions If a customer purchases a new stereo system during the spring sale, he may choose any 6 CDs from 20 classical and 30 jazz selections. In how many ways can the customer choose 3 of each?

71. Guessing on matching questions Ten words are to be paired with the correct 10 out of 12 possible defini-tions. How many ways are there of guessing?

72. Guessing on true-false exams How many possible ways are there of guessing on a 10-question true-false exam, if it is known that the instructor will have 5 true and 5 false responses?

73. Number of Wendy’s® hamburgers Wendy’s® Old Fashioned Hamburgers offers eight toppings for their single hamburger. How many different single hamburgers can be ordered?

74. Number of ice cream sundaes A restaurant offers ten toppings for their ice cream sundaes. How many different sundaes can be ordered?

PracticeUse Pascal’s Triangle to compute each combination. 75. C 18, 32 76. C 17, 42

discovery and Writing 77. Prove that C 1n, n2 5 1.

78. Prove that C 1n, 02 5 1.

79. Prove that anrb 5 a n

n 2 rb.

80. Show that the Binomial Theorem can be expressed in the form

1a 1 b2 n 5 an

k50an

kban2kbk

81. Explain how to use Pascal’s Triangle to find C 18, 52 . 82. Explain how to use Pascal’s Triangle to find C 110, 82 .

reviewFind the value of x.

83. logx 16 5 4 84. logp x 512

85. log"7 49 5 x 86. logx 12

5 213

Determine whether the statement is true or false. 87. log17 1 5 0 88. log5 0 5 1

89. logb bb 5 b

90. log7 Alog7 B

5 log7 AB

Not For Sale

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