6.1 Sets and Counting

Chapter Overview

In this chapter, you will learn to:

Use set theory and Venn diagrams to solve counting problems.
Use the Multiplication Axiom to solve counting problems.
Use Permutations to solve counting problems.
Use Combinations to solve counting problems.
Use the Binomial Theorem to expand ${(x + y)}^{n}$ .

Sets

In this section, we will familiarize ourselves with set operations and notations, so that we can apply these concepts to both counting and probability problems. We begin by defining some terms.

A set is a collection of objects, and its members are called the elements of the set. We name the set by using capital letters, and enclose its members in braces. Suppose we need to list the members of the chess club. We use the following set notation.

$C = {Ken, Bob, Tran, Shanti, Eric}$

A set that has no members is called an empty set. The empty set is denoted by the symbol Ø.

Two sets are equal if they have the same elements.

A set $A$ is a subset of a set $B$ if every member of $A$ is also a member of $B$ .

Suppose $C = {Al, Bob, Chris, David, Ed}$ and $A = {Bob, David}$ . Then $A$ is a subset of $C$ , written as $A ⊆ C$ .

Every set is a subset of itself, and the empty set is a subset of every set.

Union Of Two Sets Let $A$ and $B$ be two sets, then the union of $A$ and $B$ , written as $A \cup B$ , is the set of all elements that are either in $A$ or in $B$ , or in both $A$ and $B$ .

Intersection Of Two Sets Let $A$ and $B$ be two sets, then the intersection of $A$ and $B$ , written as $A \cap B$ , is the set of all elements that are common to both sets $A$ and $B$ .

A universal set $U$ is the set consisting of all elements under consideration.

Complement of a Set Let $A$ be any set, then the complement of set $A$ , written as $\overset{ˉ}{A}$ , is the set consisting of elements in the universal set $U$ that are not in $A$ .

Disjoint Sets Two sets $A$ and $B$ are called disjoint sets if their intersection is an empty set.

Example 4

Let the universal set $U = {red, orange, yellow, green, blue, indigo, violet}$ , and $P = {red, yellow, blue}$ . Find the complement of $P$ .

The complement of a set $P$ is the set consisting of elements in the universal set $U$ that are not in $P$ . Therefore,

$\overset{ˉ}{P} = {orange, green, indigo, violet}$

To achieve a better understanding, let us suppose that the universal set $U$ represents the colors of the spectrum, and $P$ the primary colors, then $\overset{ˉ}{P}$ represents those colors of the spectrum that are not primary colors.

Example 5

Let $U = {red, orange, yellow, green, blue, indigo, violet}$ , $P = {red, yellow, blue}$ , $Q = {red, green}$ , and $R = {orange, green, indigo}$ . Find $\bar{P \cup Q} \cap \overset{ˉ}{R}$ .

We do the problems in steps.

$\begin{matrix} P \cup Q = {red, yellow, blue, green} \\ \bar{P \cup Q} = {orange, indigo, violet} \\ \overset{ˉ}{R} = {red, yellow, blue, violet} \\ \bar{P \cup Q} \cap \overset{ˉ}{R} = {violet} \end{matrix}$

We now use Venn diagrams to illustrate the relations between sets. In the late 1800s, an English logician named John Venn developed a method to represent relationship between sets. He represented these relationships using diagrams, which are now known as Venn diagrams. A Venn diagram represents a set as the interior of a circle. Often two or more circles are enclosed in a rectangle where the rectangle represents the universal set. To visualize an intersection or union of a set is easy. In this section, we will mainly use Venn diagrams to sort various populations and count objects.

Example 6

Suppose a survey of car enthusiasts showed that over a certain time period, 30 drove cars with automatic transmissions, 20 drove cars with standard transmissions, and 12 drove cars of both types. If every one in the survey drove cars with one of these transmissions, how many people participated in the survey?

We will use Venn diagrams to solve this problem.

Let the set $A$ represent those car enthusiasts who drove cars with automatic transmissions, and set $S$ represent the car enthusiasts who drove the cars with standard transmissions. Now we use Venn diagrams to sort out the information given in this problem.

Since 12 people drove both cars, we place the number 12 in the region common to both sets.

Figure 6.1

Because 30 people drove cars with automatic transmissions, the circle A must contain 30 elements. This means $x + 12 = 30$ , or $x equals 18$ . Similarly, since 20 people drove cars with standard transmissions, the circle B must contain 20 elements, or $y + 12 = 20$ which in turn makes $y equals 8$ .

Now that all the information is sorted out, it is easy to read from the diagram that 18 people drove cars with automatic transmissions only, 12 people drove both types of cars, and 8 drove cars with standard transmissions only. Therefore, $18 + 12 + 8 = 38$ people took part in the survey.

Example 7

A survey of 100 people in California indicates that 60 people have visited Disneyland, 15 have visited Knott's Berry Farm, and 6 have visited both. How many people have visited neither place?

The problem is similar to the one in Example 6.

Let the set $D$ represent the people who have visited Disneyland, and $K$ the set of people who have visited Knott's Berry Farm.

Figure 6.2

We fill the three regions associated with the sets D and K in the same manner as before. Since 100 people participated in the survey, the rectangle representing the universal set U must contain 100 objects. Let x represent those people in the universal set that are neither in the set D nor in K. This means $54 + 6 + 9 + x = 100$ , or $x equals 31$ .

Therefore, there are 31 people in the survey who have visited neither place.

Example 8

A survey of 100 exercise conscious people resulted in the following information:

50 jog, 30 swim, and 35 cycle
14 jog and swim
7 swim and cycle
9 jog and cycle
3 people take part in all three activities

How many jog but do not swim or cycle?
How many take part in only one of the activities?
How many do not take part in any of these activities?

Let $J$ represent the set of people who jog, $S$ the set of people who swim, and $C$ who cycle.

In using Venn diagrams, our ultimate aim is to assign a number to each region. We always begin by first assigning the number to the innermost region and then working our way out.

Figure 6.3

We place a 3 in the innermost region of because it represents the number of people who participate in all three activities. Next we compute x, y and z.

Since 14 people jog and swim, $x + 3 = 14$ , or $x equals 11$ .

The fact that 9 people jog and cycle results in $y + 3 = 9$ , or $y equals 6$ .

Since 7 people swim and cycle, $z + 3 = 7$ , or $z equals 4$ .

This information is depicted.

Now we proceed to find the unknowns m, n and p.

Since 50 people jog, $m + 11 + 6 + 3 = 50$ , or $m equals 30$ .

Thirty people swim, therefore, $n + 11 + 4 + 3 = 30$ , or $n equals 12$ .

Thirty five people cycle, therefore, $p + 6 + 4 + 3 = 35$ , or $p equals 22$ .

By adding all the entries in all three sets, we get a sum of 88. Since 100 people were surveyed, the number inside the universal set but outside of all three sets is 100 – 88, or 12.

In, the information is sorted out, and the questions can readily be answered.

The number of people who jog but do not swim or cycle is 30.
The number who take part in only one of these activities is $30 + 12 + 22 = 64$ .
The number of people who do not take part in any of these activities is 12.

Tree Diagrams and the Multiplication Axiom

In this chapter, we are trying to develop counting techniques that will be used in the the related section to study probability. One of the most fundamental of such techniques is called the Multiplication Axiom. Before we introduce the multiplication axiom, we first look at some examples.

Example 9

If a woman has two blouses and three skirts, how many different outfits consisting of a blouse and a skirt can she wear?

Suppose we call the blouses $b sub 1 and b sub 2, and skirts s sub 1, s sub 2, and s sub 3$ .

We can have the following six outfits.

$b sub 1 s sub 1, b sub 1 s sub 2, b sub 1 s sub 3, b sub 2 s sub 1, b sub 2 s sub 2, b sub 2 s sub 3$

Alternatively, we can draw a tree diagram:

Figure 6.4

The tree diagram gives us all six possibilities. The method involves two steps. First the woman chooses a blouse. She has two choices: blouse one or blouse two. If she chooses blouse one, she has three skirts to match it with; skirt one, skirt two, or skirt three. Similarly if she chooses blouse two, she can match it with each of the three skirts, again. The tree diagram helps us visualize these possibilities.

The reader should note that the process involves two steps. For the first step of choosing a blouse, there are two choices, and for each choice of a blouse, there are three choices of choosing a skirt. So altogether there are $2 \cdot 3 = 6$ possibilities.

If, in the above example, we add the shoes to the outfit, we have the following problem.

Example 10

If a woman has two blouses, three skirts, and two pumps, how many different outfits consisting of a blouse, a skirt, and a pair of pumps can she wear?

Suppose we call the blouses $b sub 1$ and $b sub 2$ , the skirts $s sub 1$ , $s sub 2$ , and $s sub 3$ , and the pumps $p sub 1$ , and $p sub 2$ .

The following tree diagram results.

Figure 6.5

We count the number of branches in the tree, and see that there are 12 different possibilities. This time the method involves three steps. First, the woman chooses a blouse. She has two choices: blouse one or blouse two. Now suppose she chooses blouse one. This takes us to step two of the process which consists of choosing a skirt. She has three choices for a skirt, and let us suppose she chooses skirt two. Now that she has chosen a blouse and a skirt, we have moved to the third step of choosing a pair of pumps. Since she has two pairs of pumps, she has two choices for the last step. Let us suppose she chooses pumps two. She has chosen the outfit consisting of blouse one, skirt two, and pumps two, or $b sub 1 s sub 2 p sub 2$ . By looking at the different branches on the tree, one can easily see the other possibilities.

The important thing to observe here, again, is that this is a three step process. There are two choices for the first step of choosing a blouse. For each choice of a blouse, there are three choices of choosing a skirt, and for each combination of a blouse and a skirt, there are two choices of selecting a pair of pumps. All in all, we have $2 \cdot 3 \cdot 2 = 12$ different possibilities.

The tree diagrams help us visualize the different possibilities, but they are not practical when the possibilities are numerous. Besides, we are mostly interested in finding the number of elements in the set and not the actual possibilities. But once the problem is envisioned, we can solve it without a tree diagram. The two examples we just solved may have given us a clue to do just that.

Let us now try to solve Example 10 without a tree diagram. Recall that the problem involved three steps: choosing a blouse, choosing a skirt, and choosing a pair of pumps. The number of ways of choosing each are listed below.

**Table 6.1**
The Number of ways of choosing a blouse	The number of ways of choosing a skirt	The number of ways of choosing pumps
2	3	2

By multiplying these three numbers we get 12, which is what we got when we did the problem using a tree diagram.

The procedure we just employed is called the multiplication axiom.

The general multiplication axiom is not limited to just two tasks and can be used for any number of tasks.

Example 12

A truck license plate consists of a letter followed by four digits. How many such license plates are possible?

Since there are 26 letters and 10 digits, we have the following choices for each.

**Table 6.2**
Letter	Digit	Digit	Digit	Digit
26	10	10	10	10

Therefore, the number of possible license plates is $26 \cdot 10 \cdot 10 \cdot 10 \cdot 10 = 260,000$ .

Example 13

In how many different ways can a 3-question true-false test be answered?

Since there are two choices for each question, we have

**Table 6.3**
Question 1	Question 2	Question 3
2	2	2

Applying the multiplication axiom, we get $2 \cdot 2 \cdot 2 = 8$ different ways.

We list all eight possibilities below.

TTT, TTF, TFT, TFF, FTT, FTF, FFT, FFF

The reader should note that the first letter in each possibility is the answer corresponding to the first question, the second letter corresponds to the answer to the second question and so on. For example, TFF, says that the answer to the first question is given as true, and the answers to the second and third questions false.

Example 14

In how many different ways can four people be seated in a row?

Suppose we put four chairs in a row, and proceed to put four people in these seats.

There are four choices for the first chair we choose. Once a person sits down in that chair, there are only three choices for the second chair, and so on. We list as shown below.

Table 6.4

So there are altogether $4 \cdot 3 \cdot 2 \cdot 1 = 24$ different ways.

Example 15

How many three-letter word sequences can be formed using the letters ${A, B, C}$ if no letter is to be repeated?

The problem is very similar to Example 14.

Imagine a child having three building blocks labeled $A$ , $B$ , and $C$ . Suppose he puts these blocks on top of each other to make word sequences. For the first letter he has three choices, namely $A$ , $B$ , or $C$ . Let us suppose he chooses the first letter to be a $B$ , then for the second block which must go on top of the first, he has only two choices: $A$ or $C$ . And for the last letter he has only one choice. We list the choices below.

Table 6.5

Therefore, 6 different word sequences can be formed.

Finally, we'd like to illustrate this with a tree diagram.

Figure 6.6

All six possibilities are displayed in the tree diagram.

Permutations

In Example 15, we were asked to find the word sequences formed by using the letters ${A, B, C}$ if no letter is to be repeated. The tree diagram gave us the following six arrangements.

$ABC$ , $ACB$ , $BAC$ , $BCA$ , $CAB$ , and $CBA$ ,

Arrangements like these, where order is important and no element is repeated, are called permutations.

Example 17

How many three-letter word sequences can be formed using the letters ${A, B, C, D}$ ?

There are four choices for the first letter of our word, three choices for the second letter, and two choices for the third.

Table 6.6

Applying the multiplication axiom, we get $4 \cdot 3 \cdot 2 = 24$ different arrangements.

Example 18

How many permutations of the letters of the word ARTICLE have consonants in the first and last positions?

In the word ARTICLE, there are 4 consonants.

Since the first letter must be a consonant, we have four choices for the first position, and once we use up a consonant, there are only three consonants left for the last spot. We show as follows:

Table 6.7

Since there are no more restrictions, we can go ahead and make the choices for the rest of the positions.

So far we have used up 2 letters, therefore, five remain. So for the next position there are five choices, for the position after that there are four choices, and so on. We get

Table 6.8

So the total permutations are $4 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1 \cdot 3 = 1440$ .

We often encounter situations where we have a set of $n$ objects and we are selecting $r$ objects to form permutations. We refer to this as permutations of $n$ objects taken $r$ at a time, and we write it as $n Pr$ .

Therefore, Example 19 can also be answered as listed below.

The number of four-letter word sequences is $5 P 4 equals 120$ .
The number of three-letter word sequences is $5 P 3 equals 60$ .
The number of two-letter word sequences is $5 P 2 equals 20$ .

Before we give a formula for $n Pr$ , we'd like to introduce a symbol that we will use a great deal in this as well as in the related section.

Now we define $n Pr$ .

Example 21 The Number of Permutations of $n$ Objects Taken $r$ at a Time

$n Pr = n (n - 1) (n - 2) (n - 3) \dots (n - r + 1)$ , or
$n Pr = \frac{n!}{(n - r)!}$

Where $n$ and $r$ are natural numbers.

The reader should become familiar with both formulas and should feel comfortable in applying either.

Next we consider some more permutation problems to get further insight into these concepts.

Example 23

In how many different ways can 4 people be seated in a straight line if two of them insist on sitting next to each other?

Let us suppose we have four people $A$ , $B$ , $C$ , and $D$ . Further suppose that $A$ and $B$ want to sit together. For the sake of argument, we tie $A$ and $B$ together and treat them as one person.

The four people are $\begin{matrix} AB \end{matrix}$ $\begin{matrix} CD \end{matrix}$ . Since $\begin{matrix} AB \end{matrix}$ is treated as one person, we have the following possible arrangements.

$\begin{matrix} AB \end{matrix} \begin{matrix} CD \end{matrix}, \begin{matrix} AB \end{matrix} \begin{matrix} DC \end{matrix}, \begin{matrix} C \end{matrix} \begin{matrix} AB \end{matrix} \begin{matrix} D \end{matrix}, \begin{matrix} D \end{matrix} \begin{matrix} AB \end{matrix} \begin{matrix} C \end{matrix}, \begin{matrix} CD \end{matrix} \begin{matrix} AB \end{matrix}, \begin{matrix} DC \end{matrix} \begin{matrix} AB \end{matrix}$

Note that there are six more such permutations because $A$ and $B$ could also be tied in the order $BA$ . And they are

$\begin{matrix} BA \end{matrix} \begin{matrix} CD \end{matrix}, \begin{matrix} BA \end{matrix} \begin{matrix} DC \end{matrix}, \begin{matrix} C \end{matrix} \begin{matrix} BA \end{matrix} \begin{matrix} D \end{matrix}, \begin{matrix} D \end{matrix} \begin{matrix} BA \end{matrix} \begin{matrix} C \end{matrix}, \begin{matrix} CD \end{matrix} \begin{matrix} BA \end{matrix}, \begin{matrix} DC \end{matrix} \begin{matrix} BA \end{matrix}$

So altogether there are 12 different permutations.

Let us now do the problem using the multiplication axiom.

After we tie two of the people together and treat them as one person, we can say we have only three people. The multiplication axiom tells us that three people can be seated in $3 !$ ways. Since two people can be tied together $2 !$ ways, there are $3 ! 2 ! equals 12$ different arrangements.

Example 24

You have 4 math books and 5 history books to put on a shelf that has 5 slots. In how many ways can the books be shelved if the first three slots are filled with math books and the next two slots are filled with history books?

We first do the problem using the multiplication axiom.

Since the math books go in the first three slots, there are 4 choices for the first slot, 3 for the second and 2 for the third. The fourth slot requires a history book, and has five choices. Once that choice is made, there are 4 history books left, and therefore, 4 choices for the last slot. The choices are shown below.

Table 6.9

Therefore, the number of permutations are $4 \cdot 3 \cdot 2 \cdot 5 \cdot 4 = 480$ .

Alternately, we can see that $4 \cdot 3 \cdot 2$ is really same as $4 P 3$ , and $5 times 4$ is $5 P 2$ .

So the answer can be written as $open parenthesis P 3 close parenthesis open parenthesis 5 P 2 close parenthesis equals 480$ .

Clearly, this makes sense. For every permutation of three math books placed in the first three slots, there are $5 P 2$ permutations of history books that can be placed in the last two slots. Hence the multiplication axiom applies, and we have the answer $open parenthesis 4 P 3 close parenthesis open parenthesis 5 P 2 close parenthesis$ .

We summarize.

Permutations
A permutation of a set of elements is an ordered arrangement where each element is used once.
Factorial
$n! = n (n - 1) (n - 2) (n - 3) \dots 3 \cdot 2 \cdot 1$ .
Where $n$ is a natural number.
$0 ! equals 1$
Permutations of $n$ Objects Taken $r$ at a Time
$n Pr = n (n - 1) (n - 2) (n - 3) \dots (n - r + 1)$ , or $n Pr = \frac{n!}{(n - r)!}$
Where $n$ and $r$ are natural numbers.

Circular Permutations and Permutations with Similar Elements

Section Overview

In this section we will address the following two problems.

In how many different ways can five people be seated in a circle?
In how many different ways can the letters of the word MISSISSIPPI be arranged?

The first problem comes under the category of Circular Permutations, and the second under Permutations with Similar Elements.

Circular Permutations Suppose we have three people named $A$ , $B$ , and $C$ . We have already determined that they can be seated in a straight line in $3 !$ or 6 ways. Our next problem is to see how many ways these people can be seated in a circle. We draw a diagram.

This figure shows how the people named A, B, and C can sit in a circle in different ways. — **Figure 6.7**

It happens that there are only two ways we can seat three people in a circle. This kind of permutation is called a circular permutation. In such cases, no matter where the first person sits, the permutation is not affected. Each person can shift as many places as they like, and the permutation will not be changed. Imagine the people on a merry-go-round; the rotation of the permutation does not generate a new permutation. So in circular permutations, the first person is considered a place holder, and where he sits does not matter.

Example 26

In how many different ways can five people be seated at a circular table?

We have already determined that the first person is just a place holder. Therefore, there is only one choice for the first spot. We have

Table 6.10

So the answer is 24.

Example 27

In how many ways can four couples be seated at a round table if the men and women want to sit alternately?

We again emphasize that the first person can sit anywhere without affecting the permutation.

So there is only one choice for the first spot. Suppose a man sat down first. The chair next to it must belong to a woman, and there are 4 choices. The next chair belongs to a man, so there are three choices and so on. We list the choices below.

Table 6.11

So the answer is 144.

Now we address the second problem.

Permutations with Similar Elements

Let us determine the number of distinguishable permutations of the letters ELEMENT.

Suppose we make all the letters different by labeling the letters as follows.

$E sub 1 LE sub 2 ME sub 3 NT$

Since all the letters are now different, there are $7 !$ different permutations.

Let us now look at one such permutation, say

$LE sub 1 ME sub 2 NE sub 3 T$

Suppose we form new permutations from this arrangement by only moving the E's. Clearly, there are $3 !$ or 6 such arrangements. We list them below.

$LE sub 1 ME sub 2 NE sub 3 T$

$LE sub 1 ME sub 3 NE sub 2 T$

$LE sub 2 ME sub 1 NE sub 3 T$

$LE sub 3 ME sub 3 NE sub 1 T$

$LE sub 3 ME sub 2 NE sub 1 T$

$LE sub 3 ME sub 1 NE sub 2 T$

Because the $E$ 's are not different, there is only one arrangement $LEMENET$ and not six. This is true for every permutation.

Let us suppose there are $n$ different permutations of the letters $ELEMENT$ .

Then there are $n times 3 !$ permutations of the letters $E sub 1 LE sub 2 ME sub 3 NT$ .

But we know there are $7 !$ permutations of the letters $E sub 1 LE sub 2 ME sub 3 NT$ .

Therefore, $n \cdot 3! = 7!$

Or $n = \frac{7!}{3!}$ .

This gives us the method we are looking for.

Example 28 Permutations with Similar Elements

The number of permutations of $n$ elements taken $n$ at a time, with $r sub 1$ elements of one kind, $r sub 2$ elements of another kind, and so on, is

$\frac{n!}{r_{1}! r_{2}! . . . r_{k}!}$

We summarize.

Example 33

Circular Permutations
The number of permutations of $n$ elements in a circle is
$open parenthesis n minus 1 close parenthesis !$
Permutations with Similar Elements
The number of permutations of $n$ elements taken $n$ at a time, with $r sub 1$ elements of one kind, $r sub 2$ elements of another kind, and so on, such that $n = r_{1} + r_{2} + \dots + r_{k}$ is
$\frac{n!}{r_{1}! r_{2}! \dots r_{k}!}$
This is also referred to as ordered partitions.

Combinations

Suppose we have a set of three letters ${A, B, C}$ , and we are asked to make two-letter word sequences. We have the following six permutations.

$AB$ $BA$ $BC$ $CB$ $AC$ $CA$

Now suppose we have a group of three people ${A, B, C}$ as Al, Bob, and Chris, respectively, and we are asked to form committees of two people each. This time we have only three committees, namely,

$AB$ $BC$ $AC$

When forming committees, the order is not important, because the committee that has Al and Bob is no different than the committee that has Bob and Al. As a result, we have only three committees and not six.

Forming word sequences is an example of permutations, while forming committees is an example of combinations – the topic of this section.

Permutations are those arrangements where order is important, while combinations are those arrangements where order is not significant. From now on, this is how we will tell permutations and combinations apart.

In Example 32, there were six permutations, but only three combinations.

Just as the symbol $n Pr$ represents the number of permutations of $n$ objects taken $r$ at a time, $nCr$ represents the number of combinations of $n$ objects taken $r$ at a time.

So in Example 32, $3 P 2 equals 6$ , and $3 C 2 equals 3$ .

Our next goal is to determine the relationship between the number of combinations and the number of permutations in a given situation.

In Example 32, if we knew that there were three combinations, we could have found the number of permutations by multiplying this number by $2 !$ . That is because each combination consists of two letters, and that makes $2 !$ permutations.

Example 34

Given the set of letters ${A, B, C, D}$ . Write the number of combinations of three letters, and then from these combinations determine the number of permutations.

We have the following four combinations.

ABC BCD CDA BDA

Since every combination has three letters, there are $3 !$ permutations for every combination. We list them below.

ABC BCD CDA BDA

ACB BDC CAD BAD

BAC CDB DAC DAB

BCA CBD DCA DBA

CAB DCB ACD ADB

CBA DBC ADC ABD

The number of permutations are $3 !$ times the number of combinations. That is

$4 p 3 = 3! \cdot 4 C 3$

or $4 C 3 = \frac{4}{P}$

In general, $nCr = \frac{n Pr}{r!}$

Since $n Pr = \frac{n!}{(n - r)!}$

We have, $nCr = \frac{n!}{(n - r)! r!}$

Summarizing,

Example 35

Combinations
A combination of a set of elements is an arrangement where each element is used once, and order is not important.
The Number of Combinations of $n$ Objects Taken $r$ at a Time
$nCr = \frac{n!}{(n - r)! r!}$
Where $n$ and $r$ are natural numbers.

Example 40

The shopping area of a town is in the shape of square that is 5 blocks by 5 blocks. How many different routes can a taxi driver take to go from one corner of the shopping area to the opposite cater-corner?

Let us suppose the taxi driver drives from the point $A$ , the lower left hand corner, to the point $B$ , the upper right hand corner as shown in the figure below.

Figure 6.8

To reach his destination, he has to travel ten blocks; five horizontal, and five vertical. So if out of the ten blocks he chooses any five horizontal, the other five will have to be the vertical blocks, and vice versa. Therefore, all he has to do is to choose 5 out of ten.

The answer is 10C5, or 252.

Alternately, the problem can be solved by permutations with similar elements.

The taxi driver's route consists of five horizontal and five vertical blocks. If we call a horizontal block $H$ , and a vertical block a $V$ , then one possible route may be as follows.

$HHHHHVVVVV$

Clearly there are $\frac{10!}{5! 5!} = 252 permutations$ .

Further note that by definition $10 C5 = \frac{10!}{5! 5!}$ .

Example 41

If a coin is tossed six times, in how many ways can it fall four heads and two tails?

First we solve this problem using technique–permutations with similar elements.

We need 4 heads and 2 tails, that is

$HHHHTT$

There are $\frac{6!}{4! 2!} = 15$ permutations.

Now we solve this problem using combinations.

Suppose we have six spots to put the coins on. If we choose any four spots for heads, the other two will automatically be tails. So the problem is simply

$6 C 4 equals 15$ .

Incidentally, we could have easily chosen the two tails, instead. In that case, we would have gotten

$6 C 2 equals 15$ .

Further observe that by definition

$6 C 4 = \frac{6!}{2! 4!}$

and $6 C 2 = \frac{6!}{4! 2!}$

Which implies

$6 C 4 equals 6 C 2.$

Combinations: Involving Several Sets

So far we have solved the basic combination problem of $r$ objects chosen from $n$ different objects. Now we will consider certain variations of this problem.

Example 42

How many five-people committees consisting of 2 men and 3 women can be chosen from a total of 4 men and 4 women?

We list 4 men and 4 women as follows:

$M sub 1 M sub 2 M sub 3 M sub 4 W sub 1 W sub 2 W sub 3 W sub 4$

Since we want 5-people committees consisting of 2 men and 3 women, we'll first form all possible two-man committees and all possible three-woman committees. Clearly there are 4C2 = 6 two-man committees, and 4C3 = 4 three-woman committees, we list them as follows:

**Table 6.12**
2-Man Committees	3-Woman Committees
$M sub 1 M sub 2$	$W sub 1 W sub 2 W sub 3$
$M sub 1 M sub 3$	$W sub 1 W sub 2 W sub 4$
$M sub 1 M sub 4$	$W sub 1 W sub 3 W sub 4$
$M sub 2 M sub 3$	$W sub 2 W sub 3 W sub 4$
$M sub 2 M sub 4$
$M sub 3 M sub 4$

For every 2-man committee there are four 3-woman committees that can be chosen to make a 5-person committee. If we choose $M sub 1 M sub 2$ as our 2-man committee, then we can choose any of $W sub 1 W sub 2 W sub 3$ , $W sub 1 W sub 2 W sub 4$ , $W sub 1 W sub 3 W sub 4$ , or $W sub 2 W sub 3 W sub 4$ as our 3-woman committees. As a result, we get

$M sub 1 M sub 2, W sub 1 W sub 2 W sub 3 M sub 1 M sub 2, W sub 1 W sub 2 W sub 4 M sub 1 M sub 2, W sub 1 W sub 3 W sub 4 M sub 1 M sub 2, W sub 2 W sub 3 W sub 4$

Similarly, if we choose $M sub 1 M sub 3$ as our 2-man committee, then, again, we can choose any of $W sub 1 W sub 2 W sub 3$ , $W sub 1 W sub 2 W sub 4$ , $W sub 1 W sub 3 W sub 4$ , or $W sub 2 W sub 3 W sub 4$ as our 3-woman committees.

$M sub 1 M sub 3, W sub 1 W sub 2 W sub 3 M sub 1 M sub 3, W sub 1 W sub 2 W sub 4 M sub 1 M sub 3, W sub 1 W sub 3 W sub 4 M sub 1 M sub 3, W sub 2 W sub 3 W sub 4$

And so on.

Since there are six 2-man committees, and for every 2-man committee there are four 3- woman committees, there are altogether $6 \cdot 4 = 24$ five-people committees.

In essence, we are applying the multiplication axiom to the different combinations.

Example 43

A high school club consists of 4 freshmen, 5 sophomores, 5 juniors, and 6 seniors. How many ways can a committee of 4 people be chosen that includes

One student from each class?
All juniors?
Two freshmen and 2 seniors?
No freshmen?
At least three seniors?

Applying the multiplication axiom to the combinations involved, we get
$4 C 1 \cdot 5 C 1 \cdot 5 C 1 \cdot 6 C 1 = 600$
We are choosing all 4 members from the 5 juniors, and none from the others.
$5 C 4 equals 5$
$4 C 2 \cdot 6 C 2 = 90$
Since we don't want any freshmen on the committee, we need to choose all members from the remaining 16. That is
$16 C4 equals 1820$
Of the 4 people on the committee, we want at least three seniors. This can be done in two ways. We could have three seniors, and one non-senior, or all four seniors.
$6 C 3 \cdot 14 C1 + 6 C 4 = 295$

Example 45

A standard deck of playing cards has 52 cards consisting of 4 suits each with 13 cards. In how many different ways can a 5-card hand consisting of four cards of one suit and one of another be drawn?

We will do the problem using the following steps. Step 1. Select a suit. Step 2. Select four cards from this suit. Step 3. Select another suit. Step 4. Select a card from that suit.

Applying the multiplication axiom, we have

**Table 6.13**
Ways of selecting a suit	Ways if selecting 4 cards from this suit	Ways if selecting the next suit	Ways of selecting a card from that suit
$4 C 1$	$13 C4$	$3 C 1$	$13 C1$

$4 C 1 \cdot 13 C 4 \cdot 3 C 1 \cdot 13 C 1 = 111,540 .$

Adapted from Applied Finite Mathematics by Rupinder Sekhon (De Anza College), originally published by OpenStax CNX (cnx.org, collection col10613), licensed under CC BY 3.0. Changes were made. License: CC-BY-3.0.