646 CHAPTER 11 Counting Methods

54. In how many ways can they arrange themselves sothat Aaron and Bobbette will be next to each other?

Seats available to A and B

(Hint: First answer the following series of questions,assuming these parts are to be accomplished in order.)(a) How many pairs of adjacent seats can A and B

occupy?(b) Now, given the two seats for A and B, in how

many orders can they be seated?(c) Now, how many seats are available for C?(d) Now, how many for D?(e) Now, how many for E?(f) Now, how many for F?Now multiply your six answers above.

55. In how many ways can they arrange themselves if themen and women are to alternate seats and a manmust sit on the aisle? (Hint: First answer the follow-ing series of questions.)(a) How many choices are there for the person to

occupy the first seat, next to the aisle? (It mustbe a man.)

(b) Now, how many choices of people may occupy thesecond seat from the aisle? (It must be a woman.)

(c) Now, how many for the third seat? (one of theremaining men)

(d) Now, how many for the fourth seat? (a woman)(e) Now, how many for the fifth seat? (a man)(f) Now, how many for the sixth seat? (a woman)Now multiply your six answers above.

56. In how many ways can they arrange themselves if themen and women are to alternate with either a man ora woman on the aisle? (Hint: First answer the fol-lowing series of questions.)(a) How many choices of people are there for the

aisle seat?(b) Now, how many are there for the second seat?

(This person may not be of the same sex as theperson on the aisle.)

(c) Now, how many choices are there for the thirdseat?

(d) Now, how many for the fourth seat?(e) Now, how many for the fifth seat?(f) Now, how many for the sixth seat?Now multiply your six answers above.

57. Try working Example 4 by considering digits in theorder first, then second, then third. Explain whatgoes wrong.

58. Try working Example 4 by considering digits in theorder third, then second, then first. Explain whatgoes wrong.

59. Repeat Example 4 but this time allow repeated dig-its. Does the order in which digits are consideredmatter in this case?

1 2 3 6X X

X XX

4

X

5

X XX X

Using Permutations and CombinationsEarlier we introduced factorials as a way of counting the number of arrangementsof a given set of objects. For example, the members of the club

Andy, Bill, Cathy, David, Evelyn

can arrange themselves in a row for a photograph in different ways. Wehave also used previous methods, like tree diagrams and the fundamental countingprinciple, to answer questions such as: How many ways can club N elect a president,a secretary, and a treasurer if no one person can hold more than one office? Thisagain is a matter of arrangements. The difference is that only three, rather than allfive, of the members are involved in each case. A common way to rephrase the basicquestion here is as follows: How many arrangements are there of five things takenthree at a time? The answer, by the fundamental counting principle, is The factors begin with 5 and proceed downward, just as in a factorial product, butdo not go all the way to 1. (In this example the product stops when there are threefactors.) We now generalize this idea.

5 4 3 60.

5! 120

N

11.3

11.3 Using Permutations and Combinations 647

Permutations In the context of counting problems, arrangements are oftencalled permutations; the number of permutations of n distinct things taken r at atime is written .* Since the number of objects being arranged cannot exceedthe total number available, we assume for our purposes here that . Applying thefundamental counting principle to arrangements of this type gives

Simplification of the last factor gives the following formula.

Pn, r nn 1 n 2 . . . n r 1 .

r nPn, r

The results of Example 1(a) and(b) are supported here. For anywhole number value of N,

. See the final line ofthe display.PN, 0 1

Permutations FormulaThe number of permutations, or arrangements, of n distinct thingstaken r at a time, where , is given by

P(n, r) n(n 1)(n 2) . . . (n r 1) .r n

The factors in this product begin at n and descend until the total number of factorsis r. We now see that the number of ways in which club N can elect a president, asecretary, and a treasurer is .P5, 3 5 4 3 60

E X A M P L E 1 Evaluate each permutation.

(a) Begin at 4 and use two factors.(b) Begin at 8 and use five factors.(c) Begin at 5 and use five factors.

Notice that is equal to . It is true for all whole numbers n that . (This is the number of possible arrangements of n distinct objects taken all n at

a time.)Permutations, in general, can also be related to factorials in the following way.

Permutations formula

Multiply and divide by

Definition of factorial

This gives us a formula for computing permutations in terms of factorials.

n!

n r!

n r n r 1 . . . 2 1.

nn 1 n 2 . . . n r 1 n r n r 1 . . . 2 1

n r n r 1 . . . 2 1

Pn, r nn 1 n 2 . . . n r 1

n!Pn, n 5!P5, 5

P5, 5 5 4 3 2 1 120

P8, 5 8 7 6 5 4 6720

P4, 2 4 3 12

*Alternative notations are and .PnrnPr

648 CHAPTER 11 Counting Methods

If n and r are very large numbers, a calculator with a factorial key and this for-mula will save a lot of work when finding permutations. See the margin note.

This screen supports the results ofExample 2.

Factorial Formula for PermutationsThe number of permutations, or arrangements, of n distinct thingstaken r at a time, where , can be calculated as

P(n, r) n!(n r)! .

r n

E X A M P L E 2 Use the factorial formula for permutations to evaluate each expression.

(a)

(b)

(c)

Concerning part (c), many calculators will not display this many digits, so youmay obtain an answer such as .

Problem Solving

Permutations can be used any time we need to know the number of size-rarrangements that can be selected from a size-n set. The word arrangement im-plies an ordering, so we use permutations only in cases where the order of theitems is important. Also, permutations apply only in cases where no repetition ofitems occurs. We summarize with the following guidelines.

8.8921857 1012

P18, 12 18!

18 12!

18!6!

8,892,185,702,400

P35, 0 35!35 0!

35!35! 1

P9, 4 9!

9 4!

9!5! 3024

Guidelines for When to Use PermutationsPermutations are applied only when

1. repetitions are not allowed, and2. order is important.

E X A M P L E 3 How many nonrepeating three-digit numbers can be writtenusing digits from the set ?

Repetitions are not allowed since the numbers are to be nonrepeating. (For example, 448 is not acceptable.) Also, order is important. (For example, 476 and 746are distinct cases.) So we use permutations:

P6, 3 6 5 4 120 .

3, 4, 5, 6, 7, 8

11.3 Using Permutations and Combinations 649

The next example involves multiple parts, and hence calls for the fundamentalcounting principle, but the individual parts can be handled with permutations.

E X A M P L E 4 Suppose certain account numbers are to consist of two lettersfollowed by four digits and then three more letters, where repetitions of letters ordigits are not allowed within any of the three groups, but the last group of letters maycontain one or both of those used in the first group. How many such accounts arepossible?

The task of designing such a number consists of three parts.

1. Determine the first set of two letters.2. Determine the set of four digits.3. Determine the final set of three letters.

Each part requires an arrangement without repetitions, which is a permutation. Mul-tiply together the results of the three parts.

Part 1 Part 2 Part 3

Combinations We introduced permutations to evaluate the number of arrange-ments of n things taken r at a time, where repetitions are not allowed. The order ofthe items was important. Recall that club could elect three officers in different ways. With three-member com-mittees, on the other hand, order is not important. The committees B, D, E and E, B,D are no different. The possible number of committees is not the number of arrange-ments of size 3. Rather, it is the number of subsets of size 3 (since the order of ele-ments in a set makes no difference).

Subsets in this new context are called combinations. The number of combina-tions of n things taken r at a time (that is, the number of size r subsets, given a setof size n) is written .*

The size-3 committees (subsets) of the club (set) include:

There are ten subsets of 3, so ten is the number of three-member committees possible.Just as with permutations, repetitions are not allowed. For example, is not avalid three-member subset, just as EEB is not a valid three-member arrangement.

To see how to find the number of such subsets without listing them all, noticethat each size-3 subset (combination) gives rise to six size-3 arrangements (permu-tations). For example, the single combination ADE yields these six permutations:

E, D, A .D, E, AA, E, DE, A, DD, A, EA, D, E

E, E, B

C, D, E .B, D, E ,B, C, E ,B, C, D ,A, D, E ,A, C, E ,A, C, D ,A, B, E ,A, B, D ,A, B, C ,

A, B, C, D, EN Cn, r

P5, 3 60Andy, Bill, Cathy, David, EvelynN

51,105,600,000 650 5040 15,600

P26, 2 P10, 4 P26, 3 26 25 10 9 8 7 26 25 24

Change ringing, the English way of ringing church bells,combines mathematics and music.Bells are rung first in sequence,1,2,3, . . . . Then the sequence ispermuted (changed). On sixbells, 720 different changes(different permutations of tone)can be rung: .

Composers work out changesso that musically interesting andharmonious sequences occurregularly.

The church bells are swung bymeans of ropes attached to thewheels beside them. One ringerswings each bell, listening intentlyand watching the other ringersclosely. If one ringer gets lost andstays lost, the rhythm of theringing cannot be maintained; allthe ringers have to stop.

A ringer can spend weeks justlearning to keep a bell going andmonths learning to make the bellring in exactly the right place.Errors of second mean thattwo bells are ringing at the sametime. Even errors of secondcan be heard.

110

14

P6, 6 6!

*Alternative notations are , , and .nrC nrnCr

650 CHAPTER 11 Counting Methods

Then there must be six times as many size-3 permutations as there are size-3 combina-tions, or, in other words, one-sixth as many combinations as permutations. Therefore

The 6 appears in the denominator because there are six different ways to arrange a set of three things (since ). Generalizing from this example, r things can be arranged in different ways, so we obtain the following formula.r!

3! 3 2 1 6

C5, 3 P5, 36

5 4 3

6 10 .

Calculators can performoperations involving combinationsof n things taken r at a time.Compare this display to the resultin the text. Combinations Formula

The number of combinations, or subsets, of n distinct things taken r ata time, where , is given by

C(n, r) P(n, r)r!

n(n 1)(n 2) . . . (n r 1)r(r 1)(r 2) . . . 2 1 .

r n

Combining this formula with the factorial formula for permutations gives a wayof also evaluating combinations entirely in terms of factorials.

Cn, r Pn, r

r!

n!n r!

r!

n!r! n r!

Factorial Formula for CombinationsThe number of combinations, or subsets, of n distinct things taken r ata time, where , can be calculated as

C(n, r) P(n, r)r!

n!r!(n r)! .

r n

E X A M P L E 5 Use the factorial formula for combinations to evaluate eachexpression.

(a) (b)

Problem Solving

The following guidelines stress that combinations have an important commonproperty with permutations (repetitions are not allowed), but also an importantdistinction (order is not important with combinations). It is extremely importantto realize this when solving problems.

C22, 15 22!15! 7! 170,544C8, 6

8!6! 2!

40,320720 2

28

Bilateral cipher (above) wasinvented by Francis Bacon early inthe seventeenth century to codepolitical secrets. This binary code,a and b in combinations of five,has 32 permutations. Baconsbiformed alphabet (bottom fourrows) uses two type fonts toconceal a message in somestraight text. The decoderdeciphers a string of as and bs,groups them by fives, thendeciphers letters and words. Thiscode was applied to Shakespearesplays in efforts to prove Bacon therightful author.

11.3 Using Permutations and Combinations 651

The set of 52 playing cards in thestandard deck has four suits.

spades diamonds hearts clubs

Ace is the unit card. Jacks,queens, and kings are facecards. Each suit contains thirteendenominations: ace, 2, 3, . . . , 10,jack, queen, king. (In some games,ace rates above king, instead ofcounting as 1.)

Guidelines for When to Use CombinationsCombinations are applied only when

1. repetitions are not allowed, and2. order is not important.

E X A M P L E 6 Find the number of different subsets of size 2 in the set. List them to check the answer.

A subset of size 2 must have two distinct elements, so repetitions are not allowed. And since the order in which the elements of a set are listed makes no difference, we see that order is not important. Use the combinations formula with

and .

or, alternatively,

The six subsets of size 2 are , , , , , . c, db, db, ca, da, ca, b

C4, 2 4!

2! 4 2!

4!2! 2!

6

C4, 2 P4, 2

2!

4 32 1

6

r 2n 4

a, b, c, d

E X A M P L E 7 A common form of poker involves hands (sets) of five cardseach, dealt from a standard deck consisting of 52 different cards (illustrated in themargin). How many different 5-card hands are possible?

A 5-card hand must contain five distinct cards, so repetitions are not allowed.Also, the order is not important since a given hand depends only on the cards it con-tains, and not on the order in which they were dealt or the order in which they arelisted. Since order does not matter, use combinations (and a calculator):

or C52, 5 52!5! 52 5! 52!

5! 47! 2,598,960 .

C52, 5 P52, 55! 52 51 50 49 48

5 4 3 2 1 2,598,960

E X A M P L E 8 Melvin wants to buy ten different books but can afford onlyfour of them. In how many ways can he make his selection?

The four books selected must be distinct (repetitions are not allowed), and alsothe order of the four chosen has no bearing in this case, so we use combinations:

or C10, 4 10!

4! 10 4!

10!4! 6!

210 ways.

C10, 4 P10, 4

4!

10 9 8 74 3 2 1

210 ways,

652 CHAPTER 11 Counting Methods

Notice that, according to our formula for combinations,

which is the same as . In fact, Exercise 58 asks you to prove the fact that, ingeneral, for all whole numbers n and r, with ,

See the margin note as well. It indicates that is equal to .C10, 4C10, 6

Cn, r Cn, n r .

r nC10, 4

C10, 6 10!

6! 10 6!

10!6! 4!

210 ,

The 1 indicates that thestatement

is true.

The illustration above is from the1560s text Logistica, by themathematician J. Buteo. Amongother topics, the book discussesthe number of possible throws offour dice and the number ofarrangements of the cylinders of acombination lock. Note thatcombination is a misleadingname for these locks sincerepetitions are allowed, and, also,order makes a difference.

C10, 6 C10, 4

E X A M P L E 9 How many different three-member committees could club Nappoint so that exactly one woman is on the committee?

Recall that . Two members are women;three are men. Although the question mentioned only that the committee must in-clude exactly one woman, in order to complete the committee two men must be selected as well. Therefore the task of selecting the committee members consists oftwo parts:

1. Choose one woman.2. Choose two men.

One woman can be chosen in ways, and two men can bechosen in ways. Using the fundamental counting prin-ciple gives different committees. This number is small enough to check bylisting the possibilities:

E, B, D .E, A, D ,E, A, B ,C, B, D ,C, A, D ,C, A, B ,

2 3 6C3, 2 3 22 1 3

C2, 1 21 2

N Andy, Bill, Cathy, David, Evelyn

E X A M P L E 10 Every member of the Alpha Beta Gamma fraternity would liketo attend a special event this weekend, but only ten members will be allowed to attend. How many ways could the lucky ten be selected if there are a total of 48 members?

In this case, ten distinct men are required (repetitions are not allowed), and theorder of selection makes no difference, so we use combinations. We evaluate on a calculator.

Problem Solving

Many counting problems involve selecting some of the items from a given set ofitems. The particular conditions of the problem will determine which specifictechnique to use. The following guidelines may help.

C48, 10 48!

10! 38! 6,540,715,896

11.3 Using Permutations and Combinations 653

Both permutations and combinations produce the number of ways of selectingr items from n items where repetitions are not allowed. Permutations apply toarrangements (order is important), while combinations apply to subsets (order is notimportant). These similarities and differences, as well as the appropriate formulas,are summarized in the following table.

Any problem solved with permutations (or with factorials as in the previous sec-tion) can also be solved with the fundamental counting principle. In fact, both fac-torials and permutations are simply ways to characterize certain product patterns thatarise from applying the fundamental counting principle. And combinations are reallyonly permutations modified for cases where order is not important.

Guidelines for Choosing a Counting Method1. If selected items can be repeated, use the fundamental counting

principle.Example: How many four-digit numbers are there?

2. If selected items cannot be repeated, and order is important, usepermutations.Example: How many ways can three of eight people line up at

a ticket counter?

3. If selected items cannot be repeated, and order is not important,use combinations.Example: How many ways can a committee of three be

selected from a group of twelve people?

C12, 3 12 11 10

3 2 1 220

P8, 3 8 7 6 336

9 103 9000

Permutations Combinations

Number of ways of selecting r items out of n items

Repetitions are not allowed.

Order is important. Order is not important.

Arrangements of n items taken r at a time Subsets of n items taken r at a time

Clue words: arrangement, schedule, order Clue words: set, group, sample, selection

n!

r! n r!

n!n r!

nn 1 n 2 . . . n r 1

rr 1 r 2 . . . 2 1 nn 1 n 2 . . . n r 1

Cn, rPn, r

654 CHAPTER 11 Counting Methods

Most, if not all, of the exercises in this section will call for permutations and/orcombinations. And in the case of multiple-part tasks, the fundamental counting prin-ciple may also be required.

Poker Hands

In 5-card poker, played with a standard 52-carddeck, 2,598,960 different hands are possible.(See Example 7.) The desirability of the various

hands depends upon their relative chance ofoccurrence, which, in turn, depends on thenumber of different ways they can occur, asshown in Table 4.

F O R F U R T H E R T H O U G H T

TABLE 4 Categories of Hands in 5-Card Poker

Number of OutcomesEvent E Description of Event E Favorable to E

Royal flush Ace, king, queen, jack, and 10, all of the same suit 4Straight flush 5 cards of consecutive denominations, all in the 36

same suit (excluding royal flush)Four of a kind 4 cards of the same denomination, plus 1 624

additional cardFull house 3 cards of one denomination, plus 2 cards of a 3744

second denominationFlush Any 5 cards all of the same suit (excluding royal 5108

flush and straight flush)Straight 5 cards of consecutive denominations (not all the 10,200

same suit)Three of a kind 3 cards of one denomination, plus 2 cards of two 54,912

additional denominationsTwo pairs 2 cards of one denomination, plus 2 cards of a 123,552

second denomination, plus 1 card of a thirddenomination

One pair 2 cards of one denomination, plus 3 additional 1,098,240cards of three different denominations

No pair No two cards of the same denomination (and not all 1,302,540the same suit)

Total 2,598,960

For Group Discussion

As the table shows, a full house is a relativelyrare occurrence. (Only four of a kind, straightflush, and royal flush are less likely.) To verifythat there are 3744 different full house handspossible, carry out the following steps.

1. Explain why there are C(4, 3) different waysto select three aces from the deck.

2. Explain why there are C(4, 2) different waysto select two 8s from the deck.

3. If aces and 8s (three aces and two 8s) isone kind of full house, show that there areP(13, 2) different kinds of full housealtogether.

4. Multiply the expressions from Steps 1, 2,and 3 together. Explain why this productshould give the total number of full househands possible.

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