Counting Principles

Learning Objectives

Solve counting problems using the addition principle.
Solve counting problems using the multiplication principle.

Objective 1: Solve counting problems using the addition principle.

In probability theory, an outcome is a possible result of an experiment or trial.

In probability theory, an event is a set of outcomes of an experiment.

Disjoint events cannot happen at the same time. In other words, they are mutually exclusive.

The addition principle applies when we are making only one selection.

Example 1

Solve counting problems using the addition principle.

ⓐ Seven red and five green marbles are placed in a bag. How many marbles are there to choose from?

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

Solution

ⓐ There are 7 ways of picking a red marble and 5 ways of picking a green marble and we cannot pick a red and a green at the same time. Therefore, there are 7+5 = 12 ways of picking a marble.

ⓑ There are 3 negative numbers in A and 3 even numbers in A and the even numbers are not negative. Therefore, there are 3+3 = 6 ways of choosing a negative or an even number from A.

Practice Makes Perfect

Solve counting problems using the addition principle.

Ten red and six green marbles are placed in a bag. How many marbles are there to choose from?

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

A young boy is deciding on a snack for the afternoon. He is deciding among 5 different chips, 3 different fruits and 2 different vegetables. What is the total number of snack options?

Objective 2: Solve counting problems using the multiplication principle.

The Multiplication Principle applies when we are making more than one selection.

Example 2

Solve counting problems using the multiplication principle.

ⓐ Diane packed 2 skirts, 3 blouses, and 2 sweaters 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.
ⓑ A restaurant offers a lunch special that includes an entree, a main dish, and a beverage. There are 3 types of entrees, 4 main dish options, and 5 beverage choices. Find the total number of possible lunch specials.
ⓒ Next semester you are going to take one science class, one math class, one history class and one english class. According to the schedule you have 4 different science classes, 3 different math classes, 2 different history classes, and 3 different English classes to choose from. Assuming no scheduling conflicts, how many different four-course selections can you make?
ⓓ How many license plates consisting of 2 letters followed by 4 digits are possible?

Solution

ⓐ There are 2 outcomes for the skirts, 3 outcomes for the blouses and 2 outcomes for the sweaters. The total number of possible outfits is then:
$\begin{matrix} \underset{⏟}{2} & \cdot & \underset{⏟}{3} & \cdot & \underset{⏟}{2} & = 12 \\ Skirts & Blouses & Sweaters \end{matrix}$
ⓑ There are 3 outcomes for the entrees, 4 outcomes for the main dish and 5 outcomes for the beverages. The total number of lunch specials is:
$\begin{matrix} \underset{⏟}{3} & \cdot & \underset{⏟}{4} & \cdot & \underset{⏟}{5} & = 60 \\ Entrees & Main Dish & Beverages \end{matrix}$
ⓒ There are 4 outcomes for the science class, 3 outcomes for the math class, 2 outcomes for the history class, and 3 outcomes for the English class. The total number of 4-course selections is:
$\begin{matrix} \underset{⏟}{4} & \cdot & \underset{⏟}{3} & \cdot & \underset{⏟}{2} & \cdot & \underset{⏟}{3} & = 72 \\ Science & Math & English & History \end{matrix}$
ⓓ There are 26 outcomes for the 1st letter, 26 outcomes for the second letter, 10 outcomes for the first digit, 10 outcomes for the second digit, 10 outcomes for the third digit, and 10 outcomes for the 4th digit. The number of license plates is:
$\begin{matrix} \underset{⏟}{26} & \cdot & \underset{⏟}{26} & \cdot & \underset{⏟}{10} & \cdot & \underset{⏟}{10} & \cdot & \underset{⏟}{10} & \cdot & \underset{⏟}{10} & = 6, 760, 000 \\ 1 st letter & 2 nd letter & 1 st digit & 2 nd digit & 3 rd digit & 4 th digit \end{matrix}$

Practice Makes Perfect

Solve counting problems using the multiplication principle.

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

If you have three types of meat to make a sandwich (turkey, roast beef, and ham), and two types of bread (wheat and rye), how many different sandwiches with one kind of meat can be created?

Next semester you are going to take one science class, one math class, one history class and one english class. According to the schedule you have 4 different science classes, 3 different math classes, 2 different history classes, and 3 different English classes to choose from. Assuming no scheduling conflicts, how many different four-course selections can you make?

In Missouri, license plates have 3 letters and 3 numbers. How many license plates, consisting of 3 letters followed by 3 digits are possible?

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.

The addition of 3 iPods and 4 iPhones. — Figure 1

Example 3

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.

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.

A tree diagram of the different menu combinations. — Figure 2

The possible choices are:

soup, chicken, cake
soup, chicken, pudding
soup, fish, cake
soup, fish, pudding
soup, steak, cake
soup, steak, pudding
salad, chicken, cake
salad, chicken, pudding
salad, fish, cake
salad, fish, pudding
salad, steak, cake
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.

\begin{array}{l} # of appetizer options \times & # of entree options \times & # of dessert options \\ 2 \times & 3 \times & 2 & = 12 \end{array}

Example 4

Using the Multiplication Principle

Diane packed 2 skirts, 4 blouses, and 2 sweaters 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.

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.

The image shows three horizontal blank lines. An 'X' is positioned above the middle blank line and another 'X' is positioned above the rightmost blank line. The leftmost blank line does not have an 'X' above it.

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.

A mathematical equation shows '4 x 3 x 2 = 24', demonstrating the multiplication of three single-digit numbers resulting in 24.

There are 24 possible permutations of the paintings.

Example 5

Finding the Number of Permutations Using the Multiplication Principle

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

ⓐHow many ways can they place first, second, and third?
ⓑ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.)
ⓒHow many ways can all nine swimmers line up for a photo?

Solution

ⓐDraw lines for each place.

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.

Multiply to find that there are 504 ways for the swimmers to place.
ⓑDraw lines for describing each place.

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.

Multiply to find that there are 56 ways for the swimmers to place if Ariel wins first.
ⓒ
Draw lines for describing each place in the photo.

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.

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 $n$ distinct objects can always be found by $n! .$

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

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 $n$ objects and we want to choose $r$ objects from the set in order, we write $P (n, r) .$ Another way to write this is $n P_{r},$ a notation commonly seen on computers and calculators. To calculate $P (n, r),$ we begin by finding $n!,$ the number of ways to line up all $n$ objects. We then divide by $(n - r)!$ to cancel out the $(n - r)$ 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 $6 \times 5 \times 4 = 120.$ Using factorials, we get the same result.

\frac{6!}{3!} = \frac{6 \cdot 5 \cdot 4 \cdot 3!}{3!} = 6 \cdot 5 \cdot 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) = \frac{n!}{(n - r)!}

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

Example 6

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

\begin{array}{l} P (n, r) = \frac{n!}{(n - r)!} \\ P (12, 9) = \frac{12!}{(12 - 9)!} = \frac{12!}{3!} = 79, 833, 600 \end{array}

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 $_{n} P_{r}$ function, enter 9, and then press the equal sign. The $_{n} P_{r}$ function may be located under the MATH menu with probability commands.

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

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 $r$ objects from a set of $n$ objects where the order does not matter can be written as $C (n, r) .$ Just as with permutations, $C (n, r)$ can also be written as $_{n} C_{r} .$ In this case, the general formula is as follows.

C (n, r) = \frac{n!}{r! (n - r)!}

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 $3! = 3 \cdot 2 \cdot 1 = 6$ ways to order 3 paintings. There are $\frac{24}{6},$ 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.

Example 7

Finding the Number of Combinations Using the Formula

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

ⓐHow many ways can you select your side dishes?
ⓑHow many ways can you select 3 side dishes?

Solution

ⓐWe want to choose 2 side dishes from 5 options.
$C (5, 2) = \frac{5!}{2! (5 - 2)!} = 10$
ⓑWe want to choose 3 side dishes from 5 options.
$C (5, 3) = \frac{5!}{3! (5 - 3)!} = 10$

Analysis

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

Finding the Number of Subsets of a Set

We have looked only at combination problems in which we chose exactly $r$ 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 $C (5, 0) = 1$ way to order a pizza with no toppings. There are $C (5, 1) = 5$ 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

There are 32 possible pizzas. This result is equal to $2^{5} .$

We are presented with a sequence of choices. For each of the $n$ objects we have two choices: include it in the subset or not. So for the whole subset we have made $n$ choices, each with two options. So there are a total of $2 \cdot 2 \cdot 2 \cdot \dots \cdot 2$ 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.

Example 8

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 $n = 4$ into the formula.

\begin{array}{l} 2^{n} = 2^{4} \\ = 16 \end{array}

There are 16 possible ways to order a potato.

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 $12!$ 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 $12!$ permutations we counted are duplicates. The general formula for this situation is as follows.

\frac{n!}{r_{1}! r_{2}! \dots 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 $4!$ ways to order the stars and $3!$ ways to order the moon.

\frac{12!}{4! 3!} = 3, 326, 400

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

Example 9

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 $n = 8, r_{1} = 2,$ and $r_{2} = 2$ into the formula.

\frac{8!}{2! 2!} = 10, 080

There are 10,080 arrangements.

Key Equations

..
number of permutations of $n$ distinct objects taken $r$ at a time	$P (n, r) = \frac{n!}{(n - r)!}$
number of combinations of $n$ distinct objects taken $r$ at a time	$C (n, r) = \frac{n!}{r! (n - r)!}$
number of permutations of $n$ non-distinct objects	$\frac{n!}{r_{1}! r_{2}! \dots r_{k}!}$

Key Concepts

If one event can occur in $m$ ways and a second event with no common outcomes can occur in $n$ ways, then the first or second event can occur in $m + n$ ways. See Example 3.
If one event can occur in $m$ ways and a second event can occur in $n$ ways after the first event has occurred, then the two events can occur in $m \times n$ ways. See Example 4.
A permutation is an ordering of $n$ objects.
If we have a set of $n$ objects and we want to choose $r$ objects from the set in order, we write $P (n, r) .$
Permutation problems can be solved using the Multiplication Principle or the formula for $P (n, r) .$ See Example 5 and Example 6.
A selection of objects where the order does not matter is a combination.
Given $n$ distinct objects, the number of ways to select $r$ objects from the set is $C (n, r)$ and can be found using a formula. See Example 7.
A set containing $n$ distinct objects has $2^{n}$ subsets. See Example 8.
For counting problems involving non-distinct objects, we need to divide to avoid counting duplicate permutations. See Example 9.

Section Exercises

Verbal

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

Exercise 1

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

Solution

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

Exercise 2

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

Answer the following questions.

Exercise 3

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 4

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

Exercise 5

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

Solution

A combination; $C (n, r) = \frac{n!}{(n - r)! r!}$

Numeric

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

Exercise 6

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

Exercise 7

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

Solution

$4 + 2 = 6$

Exercise 8

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

Exercise 9

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

$5 + 4 + 7 = 16$

Exercise 10

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

Exercise 11

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

Solution

$2 \times 6 = 12$

Exercise 12

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

Exercise 13

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

Solution

$10^{3} = 1000$

Exercise 14

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 15

$P (5, 2)$

Solution

$P (5, 2) = 20$

Exercise 16

$P (8, 4)$

Exercise 17

$P (3, 3)$

Solution

$P (3, 3) = 6$

Exercise 18

$P (9, 6)$

Exercise 19

$P (11, 5)$

Solution

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

Exercise 20

$C (8, 5)$

Exercise 21

$C (12, 4)$

Solution

$C (12, 4) = 495$

Exercise 22

$C (26, 3)$

Exercise 23

$C (7, 6)$

Solution

$C (7, 6) = 7$

Exercise 24

$C (10, 3)$

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

Exercise 25

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

Solution

$2^{10} = 1024$

Exercise 26

${a, b, c, \dots, z}$

Exercise 27

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

Solution

$2^{12} = 4096$

Exercise 28

The set of even numbers from 2 to 28

Exercise 29

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

Solution

$2^{9} = 512$

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

Exercise 30

The letters in the word “juggernaut”

Exercise 31

The letters in the word “academia”

Solution

$\frac{8!}{3!} = 6720$

Exercise 32

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

Exercise 33

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

Solution

$\frac{12!}{3! 2! 3! 4!}$

Exercise 34

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

Extensions

Exercise 35

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

Solution

Exercise 36

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

Exercise 37

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

Solution

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

Exercise 38

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

Exercise 39

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

Solution

$\frac{6!}{2!} \times 4! = 8640$

Real-World Applications

Exercise 40

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.

ⓐHow many arrangements are possible with no restrictions?
ⓑHow many arrangements are possible if the parents must sit in the front?

Exercise 41

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

$6 - 3 + 8 - 3 = 8$

Exercise 42

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 43

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

$4 \times 2 \times 5 = 40$

Exercise 44

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 45

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

$4 \times 12 \times 3 = 144$

Exercise 46

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

Exercise 47

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

Solution

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

Exercise 48

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 49

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

$C (10, 3) \times C (6, 5) \times C (5, 2) = 7, 200$

Exercise 50

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 51

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

$2^{11} = 2048$

Exercise 52

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 53

Suni 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

$\frac{20!}{6! 6! 8!} = 116, 396, 280$

Exercise 54

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

Addition Principle

if one event can occur in

m

ways and a second event with no common outcomes can occur in

n

ways, then the first or second event can occur in

m + n

ways

combination

a selection of objects in which order does not matter

Fundamental Counting Principle

if one event can occur in

m

ways and a second event can occur in

n

ways after the first event has occurred, then the two events can occur in

m \times n

ways; also known as the Multiplication Principle

Multiplication Principle

if one event can occur in

m

ways and a second event can occur in

n

ways after the first event has occurred, then the two events can occur in

m \times n

ways; also known as the Fundamental Counting Principle

permutation

a selection of objects in which order matters