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Counting Principles

Module by: First Last. E-mail the author

Summary: In this section, you will:

  • Solve counting problems using the Addition Principle.
  • Solve counting problems using the Multiplication Principle.
  • Solve counting problems using permutations involving n distinct objects.
  • Solve counting problems using combinations.
  • Find the number of subsets of a given set.
  • Solve counting problems using permutations involving n non-distinct objects.

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A new company sells customizable cases for tablets and smartphones. Each case comes in a variety of colors and can be personalized for an additional fee with images or a monogram. A customer can choose not to personalize or could choose to have one, two, or three images or a monogram. The customer can choose the order of the images and the letters in the monogram. The company is working with an agency to develop a marketing campaign with a focus on the huge number of options they offer. Counting the possibilities is challenging!

We encounter a wide variety of counting problems every day. There is a branch of mathematics devoted to the study of counting problems such as this one. Other applications of counting include secure passwords, horse racing outcomes, and college scheduling choices. We will examine this type of mathematics in this section.

Using the Addition Principle

The company that sells customizable cases offers cases for tablets and smartphones. There are 3 supported tablet models and 5 supported smartphone models. The Addition Principle tells us that we can add the number of tablet options to the number of smartphone options to find the total number of options. By the Addition Principle, there are 8 total options, as we can see in Figure 1.

Figure 1
The addition of 3 iPods and 4 iPhones.

A General Note: The Addition Principle:

According to the Addition Principle, if one event can occur in m m ways and a second event with no common outcomes can occur in n n ways, then the first or second event can occur in m+n m+n ways.

Example 1

Problem 1

Using the Addition Principle

There are 2 vegetarian entrée options and 5 meat entrée options on a dinner menu. What is the total number of entrée options?

Solution

We can add the number of vegetarian options to the number of meat options to find the total number of entrée options.

The addition of the type of options for an entree.

There are 7 total options.

Try It:

Exercise 1

A student is shopping for a new computer. He is deciding among 3 desktop computers and 4 laptop computers. What is the total number of computer options?

Solution

7

Using the Multiplication Principle

The Multiplication Principle applies when we are making more than one selection. Suppose we are choosing an appetizer, an entrée, and a dessert. If there are 2 appetizer options, 3 entrée options, and 2 dessert options on a fixed-price dinner menu, there are a total of 12 possible choices of one each as shown in the tree diagram in Figure 2.

Figure 2
A tree diagram of the different menu combinations.

The possible choices are:

  1. soup, chicken, cake
  2. soup, chicken, pudding
  3. soup, fish, cake
  4. soup, fish, pudding
  5. soup, steak, cake
  6. soup, steak, pudding
  7. salad, chicken, cake
  8. salad, chicken, pudding
  9. salad, fish, cake
  10. salad, fish, pudding
  11. salad, steak, cake
  12. salad, steak, pudding We can also find the total number of possible dinners by multiplying.

We could also conclude that there are 12 possible dinner choices simply by applying the Multiplication Principle.

# of appetizer options × # of entree options × # of dessert options                2                   ×               3               ×                2 =12 # of appetizer options × # of entree options × # of dessert options                2                   ×               3               ×                2 =12

A General Note: The Multiplication Principle:

According to the Multiplication Principle, if one event can occur in m m ways and a second event can occur in n n ways after the first event has occurred, then the two events can occur in m×n m×n ways. This is also known as the Fundamental Counting Principle.

Example 2

Problem 1

Using the Multiplication Principle

Diane packed 2 skirts, 4 blouses, and a sweater for her business trip. She will need to choose a skirt and a blouse for each outfit and decide whether to wear the sweater. Use the Multiplication Principle to find the total number of possible outfits.

Solution

To find the total number of outfits, find the product of the number of skirt options, the number of blouse options, and the number of sweater options.

The multiplication of number of skirt options (2) times the number of blouse options (4) times the number of sweater options (2) which equals 16.

There are 16 possible outfits.

Try It:

Exercise 2

A restaurant offers a breakfast special that includes a breakfast sandwich, a side dish, and a beverage. There are 3 types of breakfast sandwiches, 4 side dish options, and 5 beverage choices. Find the total number of possible breakfast specials.

Solution

There are 60 possible breakfast specials.

Finding the Number of Permutations of n Distinct Objects

The Multiplication Principle can be used to solve a variety of problem types. One type of problem involves placing objects in order. We arrange letters into words and digits into numbers, line up for photographs, decorate rooms, and more. An ordering of objects is called a permutation.

Finding the Number of Permutations of n Distinct Objects Using the Multiplication Principle

To solve permutation problems, it is often helpful to draw line segments for each option. That enables us to determine the number of each option so we can multiply. For instance, suppose we have four paintings, and we want to find the number of ways we can hang three of the paintings in order on the wall. We can draw three lines to represent the three places on the wall.

CNX_Precalc_Figure_11_05_005.jpg

There are four options for the first place, so we write a 4 on the first line.

Four times two blanks spots.

After the first place has been filled, there are three options for the second place so we write a 3 on the second line.

Four times three times one blank spot.

After the second place has been filled, there are two options for the third place so we write a 2 on the third line. Finally, we find the product.

CNX_Precalc_Figure_11_05_008.jpg

There are 24 possible permutations of the paintings.

How To:

Given n n distinct options, determine how many permutations there are.

  1. Determine how many options there are for the first situation.
  2. Determine how many options are left for the second situation.
  3. Continue until all of the spots are filled.
  4. Multiply the numbers together.

Example 3

Problem 1
Finding the Number of Permutations Using the Multiplication Principle

At a swimming competition, nine swimmers compete in a race.

  1. How many ways can they place first, second, and third?
  2. How many ways can they place first, second, and third if a swimmer named Ariel wins first place? (Assume there is only one contestant named Ariel.)
  3. How many ways can all nine swimmers line up for a photo?
Solution
  1. Draw lines for each place.

    CNX_Precalc_Figure_11_05_009.jpg

    There are 9 options for first place. Once someone has won first place, there are 8 remaining options for second place. Once first and second place have been won, there are 7 remaining options for third place.

    CNX_Precalc_Figure_11_05_010.jpg

    Multiply to find that there are 504 ways for the swimmers to place.

  2. Draw lines for describing each place.

    CNX_Precalc_Figure_11_05_011.jpg

    We know Ariel must win first place, so there is only 1 option for first place. There are 8 remaining options for second place, and then 7 remaining options for third place.

    CNX_Precalc_Figure_11_05_012.jpg

    Multiply to find that there are 56 ways for the swimmers to place if Ariel wins first.

  3. Draw lines for describing each place in the photo.

    CNX_Precalc_Figure_11_05_013.jpg

    There are 9 choices for the first spot, then 8 for the second, 7 for the third, 6 for the fourth, and so on until only 1 person remains for the last spot.

    CNX_Precalc_Figure_11_05_014.jpg

    There are 362,880 possible permutations for the swimmers to line up.

Analysis

Note that in part c, we found there were 9! ways for 9 people to line up. The number of permutations of n n distinct objects can always be found by n!. n!.

A family of five is having portraits taken. Use the Multiplication Principle to find the following.

Try It:

Exercise 3

How many ways can the family line up for the portrait?

Solution

120

Try It:

Exercise 4

How many ways can the photographer line up 3 family members?

Solution

60

Try It:

Exercise 5

How many ways can the family line up for the portrait if the parents are required to stand on each end?

Solution

12

Finding the Number of Permutations of n Distinct Objects Using a Formula

For some permutation problems, it is inconvenient to use the Multiplication Principle because there are so many numbers to multiply. Fortunately, we can solve these problems using a formula. Before we learn the formula, let’s look at two common notations for permutations. If we have a set of n n objects and we want to choose r r objects from the set in order, we write P(n,r). P(n,r). Another way to write this is n P r , n P r , a notation commonly seen on computers and calculators. To calculate P(n,r), P(n,r), we begin by finding n!, n!, the number of ways to line up all n n objects. We then divide by ( nr )! ( nr )! to cancel out the ( nr ) ( nr ) items that we do not wish to line up.

Let’s see how this works with a simple example. Imagine a club of six people. They need to elect a president, a vice president, and a treasurer. Six people can be elected president, any one of the five remaining people can be elected vice president, and any of the remaining four people could be elected treasurer. The number of ways this may be done is 6×5×4=120. 6×5×4=120. Using factorials, we get the same result.

6! 3! = 6·5·4·3! 3! =6·5·4=120 6! 3! = 6·5·4·3! 3! =6·5·4=120

There are 120 ways to select 3 officers in order from a club with 6 members. We refer to this as a permutation of 6 taken 3 at a time. The general formula is as follows.

P(n,r)= n! (nr)! P(n,r)= n! (nr)!

Note that the formula stills works if we are choosing all n n objects and placing them in order. In that case we would be dividing by ( nn )! ( nn )! or 0!, 0!, which we said earlier is equal to 1. So the number of permutations of n n objects taken n n at a time is n! 1 n! 1 or just n!. n!.

A General Note: Formula for Permutations of n Distinct Objects:

Given n n distinct objects, the number of ways to select r r objects from the set in order is

P(n,r)= n! (nr)! P(n,r)= n! (nr)!
(4)

How To:

Given a word problem, evaluate the possible permutations.

  1. Identify n n from the given information.
  2. Identify r r from the given information.
  3. Replace n n and r r in the formula with the given values.
  4. Evaluate.

Example 4

Problem 1
Finding the Number of Permutations Using the Formula

A professor is creating an exam of 9 questions from a test bank of 12 questions. How many ways can she select and arrange the questions?

Solution

Substitute n=12 n=12 and r=9 r=9 into the permutation formula and simplify.

  P(n,r)= n! (nr)! P(12,9)= 12! (129)! = 12! 3! =79,833,600   P(n,r)= n! (nr)! P(12,9)= 12! (129)! = 12! 3! =79,833,600

There are 79,833,600 possible permutations of exam questions!

Analysis

We can also use a calculator to find permutations. For this problem, we would enter 12, press the n P r n P r function, enter 9, and then press the equal sign. The n P r n P r function may be located under the MATH menu with probability commands.

Q&A:

Could we have solved Example 4 using the Multiplication Principle?

Yes. We could have multiplied 151413121110987654 151413121110987654 to find the same answer.

A play has a cast of 7 actors preparing to make their curtain call. Use the permutation formula to find the following.

Try It:

Exercise 6

How many ways can the 7 actors line up?

Solution

P(7,7)=5,040 P(7,7)=5,040

Try It:

Exercise 7

How many ways can 5 of the 7 actors be chosen to line up?

Solution

P(7,5)=2,520 P(7,5)=2,520

Find the Number of Combinations Using the Formula

So far, we have looked at problems asking us to put objects in order. There are many problems in which we want to select a few objects from a group of objects, but we do not care about the order. When we are selecting objects and the order does not matter, we are dealing with combinations. A selection of r r objects from a set of n n objects where the order does not matter can be written as C(n,r). C(n,r). Just as with permutations, C(n,r) C(n,r) can also be written as n C r . n C r . In this case, the general formula is as follows.

C(n,r)= n! r!(nr)! C(n,r)= n! r!(nr)!

An earlier problem considered choosing 3 of 4 possible paintings to hang on a wall. We found that there were 24 ways to select 3 of the 4 paintings in order. But what if we did not care about the order? We would expect a smaller number because selecting paintings 1, 2, 3 would be the same as selecting paintings 2, 3, 1. To find the number of ways to select 3 of the 4 paintings, disregarding the order of the paintings, divide the number of permutations by the number of ways to order 3 paintings. There are 3!=3·2·1=6 3!=3·2·1=6 ways to order 3 paintings. There are 24 6 , 24 6 , or 4 ways to select 3 of the 4 paintings. This number makes sense because every time we are selecting 3 paintings, we are not selecting 1 painting. There are 4 paintings we could choose not to select, so there are 4 ways to select 3 of the 4 paintings.

A General Note: Formula for Combinations of n Distinct Objects:

Given n n distinct objects, the number of ways to select r r objects from the set is

C(n,r)= n! r!(nr)! C(n,r)= n! r!(nr)!
(7)

How To:

Given a number of options, determine the possible number of combinations.

  1. Identify n n from the given information.
  2. Identify r r from the given information.
  3. Replace n n and r r in the formula with the given values.
  4. Evaluate.

Example 5

Problem 1

Finding the Number of Combinations Using the Formula

A fast food restaurant offers five side dish options. Your meal comes with two side dishes.

  1. How many ways can you select your side dishes?
  2. How many ways can you select 3 side dishes?
Solution
  1. We want to choose 2 side dishes from 5 options.
    C(5,2)= 5! 2!(52)! =10 C(5,2)= 5! 2!(52)! =10
  2. We want to choose 3 side dishes from 5 options.
    C(5,3)= 5! 3!(53)! =10 C(5,3)= 5! 3!(53)! =10
Analysis

We can also use a graphing calculator to find combinations. Enter 5, then press n C r , n C r , enter 3, and then press the equal sign. The n C r , n C r , function may be located under the MATH menu with probability commands.

Q&A:

Is it a coincidence that parts (a) and (b) in Example 5 have the same answers?

No. When we choose r objects from n objects, we are not choosing (nr) (nr) objects. Therefore, C(n,r)=C(n,nr). C(n,r)=C(n,nr).

Try It:

Exercise 8

An ice cream shop offers 10 flavors of ice cream. How many ways are there to choose 3 flavors for a banana split?

Solution

C(10,3)=120 C(10,3)=120

Finding the Number of Subsets of a Set

We have looked only at combination problems in which we chose exactly r r objects. In some problems, we want to consider choosing every possible number of objects. Consider, for example, a pizza restaurant that offers 5 toppings. Any number of toppings can be ordered. How many different pizzas are possible?

To answer this question, we need to consider pizzas with any number of toppings. There is C(5,0)=1 C(5,0)=1 way to order a pizza with no toppings. There are C(5,1)=5 C(5,1)=5 ways to order a pizza with exactly one topping. If we continue this process, we get

C(5,0)+C(5,1)+C(5,2)+C(5,3)+C(5,4)+C(5,5)=32 C(5,0)+C(5,1)+C(5,2)+C(5,3)+C(5,4)+C(5,5)=32

There are 32 possible pizzas. This result is equal to 2 5 . 2 5 .

We are presented with a sequence of choices. For each of the n n objects we have two choices: include it in the subset or not. So for the whole subset we have made n n choices, each with two options. So there are a total of 2·2·2··2 2·2·2··2 possible resulting subsets, all the way from the empty subset, which we obtain when we say “no” each time, to the original set itself, which we obtain when we say “yes” each time.

A General Note: Formula for the Number of Subsets of a Set :

A set containing n distinct objects has 2 n 2 n subsets.

Example 6

Problem 1

Finding the Number of Subsets of a Set

A restaurant offers butter, cheese, chives, and sour cream as toppings for a baked potato. How many different ways are there to order a potato?

Solution

We are looking for the number of subsets of a set with 4 objects. Substitute n=4 n=4 into the formula.

2 n = 2 4     =16 2 n = 2 4     =16

There are 16 possible ways to order a potato.

Try It:

Exercise 9

A sundae bar at a wedding has 6 toppings to choose from. Any number of toppings can be chosen. How many different sundaes are possible?

Solution

64 sundaes

Finding the Number of Permutations of n Non-Distinct Objects

We have studied permutations where all of the objects involved were distinct. What happens if some of the objects are indistinguishable? For example, suppose there is a sheet of 12 stickers. If all of the stickers were distinct, there would be 12!12! ways to order the stickers. However, 4 of the stickers are identical stars, and 3 are identical moons. Because all of the objects are not distinct, many of the 12! 12! permutations we counted are duplicates. The general formula for this situation is as follows.

n! r 1 ! r 2 ! r k ! n! r 1 ! r 2 ! r k !

In this example, we need to divide by the number of ways to order the 4 stars and the ways to order the 3 moons to find the number of unique permutations of the stickers. There are 4! 4! ways to order the stars and 3! 3! ways to order the moon.

12! 4!3! =3,326,400 12! 4!3! =3,326,400

There are 3,326,400 ways to order the sheet of stickers.

A General Note: Formula for Finding the Number of Permutations of n Non-Distinct Objects:

If there are n n elements in a set and r 1 r 1 are alike, r 2 r 2 are alike, r 3 r 3 are alike, and so on through r k , r k , the number of permutations can be found by

n! r 1 ! r 2 ! r k ! n! r 1 ! r 2 ! r k !
(14)

Example 7

Problem 1

Finding the Number of Permutations of n Non-Distinct Objects

Find the number of rearrangements of the letters in the word DISTINCT.

Solution

There are 8 letters. Both I and T are repeated 2 times. Substitute n=8,  r 1 =2,  n=8,  r 1 =2,  and   r 2 =2    r 2 =2  into the formula.

8! 2!2! =10,080  8! 2!2! =10,080 

There are 10,080 arrangements.

Try It:

Exercise 10

Find the number of rearrangements of the letters in the word CARRIER.

Solution

840

Media:

Access these online resources for additional instruction and practice with combinations and permutations.

Key Equations

Table 1
number of permutations of n n distinct objects taken r r at a time P(n,r)= n! (nr)! P(n,r)= n! (nr)!
number of combinations of n n distinct objects taken r r at a time C(n,r)= n! r!(nr)! C(n,r)= n! r!(nr)!
number of permutations of n n non-distinct objects n! r 1 ! r 2 ! r k ! n! r 1 ! r 2 ! r k !

Key Concepts

  • If one event can occur in m m ways and a second event with no common outcomes can occur in n n ways, then the first or second event can occur in m+n m+n ways. See Example 1.
  • If one event can occur in m m ways and a second event can occur in n n ways after the first event has occurred, then the two events can occur in m×n m×n ways. See Example 2.
  • A permutation is an ordering of n n objects.
  • If we have a set of n n objects and we want to choose r r objects from the set in order, we write P(n,r). P(n,r).
  • Permutation problems can be solved using the Multiplication Principle or the formula for P(n,r). P(n,r). See Example 3 and Example 4.
  • A selection of objects where the order does not matter is a combination.
  • Given n n distinct objects, the number of ways to select r r objects from the set is C(n,r) C(n,r) and can be found using a formula. See Example 5.
  • A set containing n n distinct objects has 2 n 2 n subsets. See Example 6.
  • For counting problems involving non-distinct objects, we need to divide to avoid counting duplicate permutations. See Example 7.

Section Exercises

Verbal

For the following exercises, assume that there are n n ways an event A A can happen, m m ways an event B B can happen, and that A and B A and B are non-overlapping.

Exercise 11

Use the Addition Principle of counting to explain how many ways event A or B A or B can occur.

Solution

There are m+n m+n ways for either event A A or event B B to occur.

Exercise 12

Use the Multiplication Principle of counting to explain how many ways event A and B A and B can occur.

Answer the following questions.

Exercise 13

When given two separate events, how do we know whether to apply the Addition Principle or the Multiplication Principle when calculating possible outcomes? What conjunctions may help to determine which operations to use?

Solution

The addition principle is applied when determining the total possible of outcomes of either event occurring. The multiplication principle is applied when determining the total possible outcomes of both events occurring. The word “or” usually implies an addition problem. The word “and” usually implies a multiplication problem.

Exercise 14

Describe how the permutation of n n objects differs from the permutation of choosing r r objects from a set of n n objects. Include how each is calculated.

Exercise 15

What is the term for the arrangement that selects r r objects from a set of n n objects when the order of the r r objects is not important? What is the formula for calculating the number of possible outcomes for this type of arrangement?

Solution

A combination; C(n,r)= n! (nr)!r! C(n,r)= n! (nr)!r!

Numeric

For the following exercises, determine whether to use the Addition Principle or the Multiplication Principle. Then perform the calculations.

Exercise 16

Let the set A={5,3,1,2,3,4,5,6}. A={5,3,1,2,3,4,5,6}. How many ways are there to choose a negative or an even number from A? A?

Exercise 17

Let the set B={23,16,7,2,20,36,48,72}. B={23,16,7,2,20,36,48,72}. How many ways are there to choose a positive or an odd number from A? A?

Solution

4+2=6 4+2=6

Exercise 18

How many ways are there to pick a red ace or a club from a standard card playing deck?

Exercise 19

How many ways are there to pick a paint color from 5 shades of green, 4 shades of blue, or 7 shades of yellow?

Solution

5+4+7=16 5+4+7=16

Exercise 20

How many outcomes are possible from tossing a pair of coins?

Exercise 21

How many outcomes are possible from tossing a coin and rolling a 6-sided die?

Solution

2×6=12 2×6=12

Exercise 22

How many two-letter strings—the first letter from A A and the second letter from B B can be formed from the sets A={b,c,d} A={b,c,d} and B={a,e,i,o,u}? B={a,e,i,o,u}?

Exercise 23

How many ways are there to construct a string of 3 digits if numbers can be repeated?

Solution

10 3 =1000 10 3 =1000

Exercise 24

How many ways are there to construct a string of 3 digits if numbers cannot be repeated?

For the following exercises, compute the value of the expression.

Exercise 25

P(5,2) P(5,2)

Solution

P(5,2)=20 P(5,2)=20

Exercise 26

P(8,4) P(8,4)

Exercise 27

P(3,3) P(3,3)

Solution

P(3,3)=6 P(3,3)=6

Exercise 28

P(9,6) P(9,6)

Exercise 29

P(11,5) P(11,5)

Solution

P(11,5)=55,440 P(11,5)=55,440

Exercise 30

C(8,5) C(8,5)

Exercise 31

C(12,4) C(12,4)

Solution

C(12,4)=495 C(12,4)=495

Exercise 32

C(26,3) C(26,3)

Exercise 33

C(7,6) C(7,6)

Solution

C(7,6)=7 C(7,6)=7

Exercise 34

C(10,3) C(10,3)

For the following exercises, find the number of subsets in each given set.

Exercise 35

{1,2,3,4,5,6,7,8,9,10} {1,2,3,4,5,6,7,8,9,10}

Solution

2 10 =1024 2 10 =1024

Exercise 36

{a,b,c,,z} {a,b,c,,z}

Exercise 37

A set containing 5 distinct numbers, 4 distinct letters, and 3 distinct symbols

Solution

2 12 =4096 2 12 =4096

Exercise 38

The set of even numbers from 2 to 28

Exercise 39

The set of two-digit numbers between 1 and 100 containing the digit 0

Solution

2 9 =512 2 9 =512

For the following exercises, find the distinct number of arrangements.

Exercise 40

The letters in the word “juggernaut”

Exercise 41

The letters in the word “academia”

Solution

8! 3! =6720 8! 3! =6720

Exercise 42

The letters in the word “academia” that begin and end in “a”

Exercise 43

The symbols in the string #,#,#,@,@,$,$,$,%,%,%,%

Solution

12! 3!2!3!4! 12! 3!2!3!4!

Exercise 44

The symbols in the string #,#,#,@,@,$,$,$,%,%,%,% that begin and end with “%”

Extensions

Exercise 45

The set, S S consists of 900,000,000 900,000,000 whole numbers, each being the same number of digits long. How many digits long is a number from S? S? (Hint: use the fact that a whole number cannot start with the digit 0.)

Solution

9

Exercise 46

The number of 5-element subsets from a set containing n n elements is equal to the number of 6-element subsets from the same set. What is the value of n? n? (Hint: the order in which the elements for the subsets are chosen is not important.)

Exercise 47

Can C(n,r) C(n,r) ever equal P(n,r)? P(n,r)? Explain.

Solution

Yes, for the trivial cases r=0 r=0 and r=1. r=1. If r=0, r=0, then C(n,r)=P(n,r)=1.  C(n,r)=P(n,r)=1.  If r=1, r=1, then r=1, r=1, C(n,r)=P(n,r)=n. C(n,r)=P(n,r)=n.

Exercise 48

Suppose a set A A has 2,048 subsets. How many distinct objects are contained in A? A?

Exercise 49

How many arrangements can be made from the letters of the word “mountains” if all the vowels must form a string?

Solution

6! 2! ×4!=8640 6! 2! ×4!=8640

Real-World Applications

Exercise 50

A family consisting of 2 parents and 3 children is to pose for a picture with 2 family members in the front and 3 in the back.

  1. How many arrangements are possible with no restrictions?
  2. How many arrangements are possible if the parents must sit in the front?
  3. How many arrangements are possible if the parents must be next to each other?

Exercise 51

A cell phone company offers 6 different voice packages and 8 different data packages. Of those, 3 packages include both voice and data. How many ways are there to choose either voice or data, but not both?

Solution

63+83=8 63+83=8

Exercise 52

In horse racing, a “trifecta” occurs when a bettor wins by selecting the first three finishers in the exact order (1st place, 2nd place, and 3rd place). How many different trifectas are possible if there are 14 horses in a race?

Exercise 53

A wholesale T-shirt company offers sizes small, medium, large, and extra-large in organic or non-organic cotton and colors white, black, gray, blue, and red. How many different T-shirts are there to choose from?

Solution

4×2×5=40 4×2×5=40

Exercise 54

Hector wants to place billboard advertisements throughout the county for his new business. How many ways can Hector choose 15 neighborhoods to advertise in if there are 30 neighborhoods in the county?

Exercise 55

An art store has 4 brands of paint pens in 12 different colors and 3 types of ink. How many paint pens are there to choose from?

Solution

4×12×3=144 4×12×3=144

Exercise 56

How many ways can a committee of 3 freshmen and 4 juniors be formed from a group of 8 8 freshmen and 11 11 juniors?

Exercise 57

How many ways can a baseball coach arrange the order of 9 batters if there are 15 players on the team?

Solution

P(15,9)=1,816,214,400 P(15,9)=1,816,214,400

Exercise 58

A conductor needs 5 cellists and 5 violinists to play at a diplomatic event. To do this, he ranks the orchestra’s 10 cellists and 16 violinists in order of musical proficiency. What is the ratio of the total cellist rankings possible to the total violinist rankings possible?

Exercise 59

A motorcycle shop has 10 choppers, 6 bobbers, and 5 café racers—different types of vintage motorcycles. How many ways can the shop choose 3 choppers, 5 bobbers, and 2 café racers for a weekend showcase?

Solution

C(10,3)×C(6,5)×C(5,2)=7,200 C(10,3)×C(6,5)×C(5,2)=7,200

Exercise 60

A skateboard shop stocks 10 types of board decks, 3 types of trucks, and 4 types of wheels. How many different skateboards can be constructed?

Exercise 61

Just-For-Kicks Sneaker Company offers an online customizing service. How many ways are there to design a custom pair of Just-For-Kicks sneakers if a customer can choose from a basic shoe up to 11 customizable options?

Solution

2 11 =2048 2 11 =2048

Exercise 62

A car wash offers the following optional services to the basic wash: clear coat wax, triple foam polish, undercarriage wash, rust inhibitor, wheel brightener, air freshener, and interior shampoo. How many washes are possible if any number of options can be added to the basic wash?

Exercise 63

Susan bought 20 plants to arrange along the border of her garden. How many distinct arrangements can she make if the plants are comprised of 6 tulips, 6 roses, and 8 daisies?

Solution

20! 6!6!8! =116,396,280 20! 6!6!8! =116,396,280

Exercise 64

How many unique ways can a string of Christmas lights be arranged from 9 red, 10 green, 6 white, and 12 gold color bulbs?

Glossary

Addition Principle:
if one event can occur in m m ways and a second event with no common outcomes can occur in n n ways, then the first or second event can occur in m+n m+n ways
combination:
a selection of objects in which order does not matter
Fundamental Counting Principle:
if one event can occur in m m ways and a second event can occur in n n ways after the first event has occurred, then the two events can occur in m×n m×n ways; also known as the Multiplication Principle
Multiplication Principle:
if one event can occur in m m ways and a second event can occur in n n ways after the first event has occurred, then the two events can occur in m×n m×n ways; also known as the Fundamental Counting Principle
permutation:
a selection of objects in which order matters

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