A state motor vehicle department would like to know how many license plate numbers are available if a license plate number consists of three letters followed by three numbers (0 through 9).

SOLUTION Using Equation (4.11), if a license plate number consists of three letters followed by three numbers, the total number of possible outcomes is(26)(26)(26)(10)(10)(10) = 17,576,000.

EXAMPLE 4.13Counting Rule 2

Suppose you roll a die twice. How many different possible outcomes can occur?

SOLUTION If a six-sided die is rolled twice, using Equation (4.10), the number of differentoutcomes is 62 = 36.

EXAMPLE 4.12Rolling a Die Twice

Suppose you toss a coin five times. What is the number of different possible outcomes (thesequences of heads and tails)?

SOLUTION If you toss a two-sided coin five times, using Equation (4.10), the number ofoutcomes is 25 = 2 * 2 * 2 * 2 * 2 = 32.

EXAMPLE 4.11Counting Rule 1

4.5 Counting Rules 1

4.5 Counting RulesIn Equation (4.1) on page 134, the probability of occurrence of an outcome was defined as thenumber of ways the outcome occurs, divided by the total number of possible outcomes. Inmany instances, there are a large number of possible outcomes, and determining the exact num-ber can be difficult. In such circumstances, rules have been developed for counting the numberof possible outcomes. This section presents five different counting rules.

COUNTING RULE 1If any one of k different mutually exclusive and collectively exhaustive events can occuron each of n trials, the number of possible outcomes is equal to

(4.10)kn

Use the POWER(k, n)worksheet function in a cellformula to compute thenumber of ways of arrangingX objects selected from nobjects. For example, theformula POWER(2, 5)computes the answer forExample 4.11.

=

The second counting rule is a more general version of the first and allows for the numberof possible events to differ from trial to trial.

COUNTING RULE 2If there are events on the first trial, events on the second trial, . . . , and events onthe nth trial, then the number of possible outcomes is

(4.11)(k1)(k2) . . . (kn)

knk2k1

Use a formula that takes theproduct of successivePOWER(k, n) functions tosolve problems related tocounting rule 2. For example,the formula POWER(26,3)*POWER(10,3) computes theanswer for Example 4.13.

=

A restaurant menu has a price-fixed complete dinner that consists of an appetizer, an entrée, abeverage, and a dessert. You have a choice of five appetizers, ten entrées, three beverages, andsix desserts. Determine the total number of possible dinners.

SOLUTION Using Equation (4.11), the total number of possible dinners is(5)(10)(3)(6) = 900.

EXAMPLE 4.14Determining theNumber of DifferentDinners

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Modifying Example 4.15, if you have six books, but there is room for only four books on theshelf, in how many ways can you arrange these books on the shelf?

SOLUTION Using Equation (4.13), the number of ordered arrangements of four booksselected from six books is equal to

nPx =n!

(n - x)!=

6!(6 - 4)!

=(6)(5)(4)(3)(2)(1)

(2)(1)= 360

EXAMPLE 4.16Counting Rule 4

If a set of six books is to be placed on a shelf, in how many ways can the six books be arranged?

SOLUTION To begin, you must realize that any of the six books could occupy the first posi-tion on the shelf. Once the first position is filled, there are five books to choose from in fillingthe second position. You continue this assignment procedure until all the positions are occu-pied. The number of ways that you can arrange six books is

n! = 6! = (6)(5)(4)(3)(2)(1) = 720

EXAMPLE 4.15Counting Rule 3

2 CHAPTER 4 Basic Probability

The third counting rule involves computing the number of ways that a set of items can bearranged in order.

COUNTING RULE 3The number of ways that all n items can be arranged in order is

(4.12)

where n! is called n factorial, and 0! is defined as 1.

n! = (n)(n - 1) . . . (1)

Use the FACT(n) worksheetfunction in a cell formula tocompute how many ways nitems can be arranged.For example, the formula=FACT(6) computes 6!

Use the PERMUT(n, x)worksheet function in a cellformula to compute thenumber of ways of arrangingx objects selected from nobjects. For example, theformula =PERMUT(6, 4)computes the answer forExample 4.16.

1On many scientific calculators,there is a button labeled nPr thatallows you to compute permutations.The symbol r is used instead of x.

In many instances you need to know the number of ways in which a subset of an entiregroup of items can be arranged in order. Each possible arrangement is called a permutation.

COUNTING RULE 4: PERMUTATIONSThe number of ways of arranging X objects selected from n objects in order is

(4.13)

where

number of objects

of objects to be arranged

factorial

P is the symbol for permutations.1

= n(n - 1) . . . (1)n! = n

x = number

n = total

nPx =n!

(n - x)!

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4.5 Counting Rules 3

In many situations, you are not interested in the order of the outcomes but only in the num-ber of ways that x items can be selected from n items, irrespective of order. Each possibleselection is called a combination.

COUNTING RULE 5: COMBINATIONSThe number of ways of selecting x objects from n objects, irrespective of order, isequal to

(4.14)

where

number of objects

of objects to be arranged

factorial

C is the symbol for combinations.2

= n(n - 1) . . . (1)n! = n

x = number

n = total

nCx =n!

x!(n - x)!

Use the COMBIN(n, x)worksheet function in a cellformula to compute thenumber of ways of arrangingx objects selected from nobjects. For example, theformula =COMBIN(6, 4)computes the answer forExample 4.17.

2On many scientific calculators,there is a button labeled nCr thatallows you to compute combinations.The symbol r is used instead of x.

If you compare this rule to counting rule 4, you see that it differs only in the inclusion of aterm X! in the denominator. When permutations were used, all of the arrangements of the Xobjects are distinguishable. With combinations, the x! possible arrangements of objects areirrelevant.

Modifying Example 4.16, if the order of the books on the shelf is irrelevant, in how many wayscan you arrange these books on the shelf?

SOLUTION Using Equation (4.14), the number of combinations of four books selected fromsix books is equal to

nCx =n!

x!(n - x)!=

6!4! (6 - 4)!

=(6)(5)(4)(3)(2)(1)

(4)(3)(2)(1)(2)(1)= 15

EXAMPLE 4.17Counting Rule 5

Problems for Section 4.5APPLYING THE CONCEPTS4.57 If there are 10 multiple-choice questions on an exam,each having three possible answers, how many differentsequences of answers are there?

4.58 A lock on a bank vault consists of three dials, eachwith 30 positions. In order for the vault to open, each of thethree dials must be in the correct position.a. How many different possible dial combinations are there

for this lock?b. What is the probability that if you randomly select a posi-

tion on each dial, you will be able to open the bank vault?c. Explain why “dial combinations” are not mathematical

combinations expressed by Equation (4.14).

4.59 a. If a coin is tossed seven times, how many differentoutcomes are possible?

b. If a die is tossed seven times, how many different out-comes are possible?

c. Discuss the differences in your answers to (a) and (b).

4.60 A particular brand of women’s jeans is available inseven different sizes, three different colors, and three differ-ent styles. How many different jeans does the store managerneed to order to have one pair of each type?

4.61 You would like to make a salad that consists oflettuce, tomato, cucumber, and peppers. You go to the super-market, intending to purchase one variety of each of these

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4 CHAPTER 4 Basic Probability

ingredients. You discover that there are eight varieties oflettuce, four varieties of tomatoes, three varieties of cucum-bers, and three varieties of peppers for sale at the supermar-ket. If you buy them all, how many different salads can youmake?

4.62 If each letter is used once, how many different four-letter “words” can be made from the letters E, L, O, and V?

4.63 In Major League Baseball, there are five teams in theEastern Division of the National League: Atlanta, Florida,New York, Philadelphia, and Washington. How many differ-ent orders of finish are there for these five teams? (Assumethat there are no ties in the standings.) Do you believe thatall these orders are equally likely? Discuss.

4.64 Referring to Problem 4.63, how many different ordersof finish are possible for the first four positions?

4.65 A gardener has six rows available in his vegetable gar-den to place tomatoes, eggplant, peppers, cucumbers, beans,and lettuce. Each vegetable will be allowed one and only onerow. How many ways are there to position these vegetablesin this garden?

4.66 The trifecta in the ninth race at the local racetrackconsists of picking the correct order of finish of the firstthree horses. If there are 12 horses entered in this race, howmany trifecta outcomes are there?

4.67 The quinella at the local racetrack consists of pickingthe horses that will place first and second in a race,irrespective of order. If eight horses are entered in a race,how many quinella combinations are there?

4.68 A student has seven books that she would like to placein her backpack. However, there is room for only four books.Regardless of the arrangement, how many ways are there ofplacing four books into the backpack?

4.69 A daily lottery is conducted in which 2 winning num-bers are selected out of 100 numbers. How many differentcombinations of winning numbers are possible?

4.70 A reading list for a course contains 20 articles. Howmany ways are there to choose 3 articles from this list?

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