# 7.4: Combinations

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Learning Objectives

In this section you will learn to:

• Count the number of combinations of $$\mathrm{r}$$ out of $$\mathrm{n}$$ items (selections without regard to arrangement).
• Use factorials to perform calculations involving combinations.

Prerequisite Skills

Before you get started, take this prerequisite quiz.

1. Calculate each expression without the use of a calculator:

a. $$4!$$

b. $$\dfrac{5!}{3!}$$

a. $$4!=24$$

b. $$\dfrac{5!}{3!}=20$$

If you missed this problem, review Section 7.3. (Note that this will open in a new window.)

2. Calculate each expression:

a. $$5P2$$

b. $$12P3$$

a. $$5P2=20$$

b. $$12P3=1320$$

If you missed this problem, review Section 7.3. (Note that this will open in a new window.)

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 the above example, there were six permutations, but only three combinations.

Just as the symbol nPr 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 the above example, 3P2 = 6, and 3C2 = 3.

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

In the above example, 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 $$\PageIndex{1}$$

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.

Solution

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.

$\begin{array}{cccc} \mathrm{ABC} & \mathrm{BCD} & \mathrm{CDA} & \mathrm{BDA} \\ \mathrm{ACB} & \mathrm{BDC} & \mathrm{CAD} & \mathrm{BAD} \\ \mathrm{BAC} & \mathrm{CDB} & \mathrm{DAC} & \mathrm{DAB} \\ \mathrm{BCA} & \mathrm{CBD} & \mathrm{DCA} & \mathrm{DBA} \\ \mathrm{CAB} & \mathrm{DCB} & \mathrm{ACD} & \mathrm{ADB} \\ \mathrm{CBA} & \mathrm{DBC} & \mathrm{ADC} & \mathrm{ABD} \end{array} \nonumber$

The number of permutations are 3! times the number of combinations; that is

4P3 = 3! $$\cdot$$ 4C3

or

$4 \mathrm{C} 3=\frac{4 \mathrm{P} 3}{3 !} \nonumber$

In general, $\mathrm{nCr}=\frac{\mathrm{nPr}}{\mathrm{r} !}\nonumber$

Since $\mathrm{nPr}=\frac{\mathrm{n} !}{(\mathrm{n}-\mathrm{r}) !}\nonumber$

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

Summarizing,

Definition: 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

$\mathrm{nCr}=\frac{\mathrm{n} !}{(\mathrm{n}-\mathrm{r}) ! \mathrm{r} !}$

where $$\mathrm{n}$$ and $$\mathrm{r}$$ are natural numbers.

Example $$\PageIndex{2}$$

Compute:

1. 5C3
2. 7C3

Solution

We use the above formula.

$5 \mathrm{C} 3=\frac{5 !}{(5-3) ! 3 !}=\frac{5 !}{2 ! 3 !}=10 \nonumber$

$7 \mathrm{C} 3=\frac{7 !}{(7-3) ! 3 !}=\frac{7 !}{4 ! 3 !}=35 \nonumber$

Example $$\PageIndex{3}$$

In how many different ways can a student select to answer five questions from a test that has seven questions, if the order of the selection is not important?

Solution

Since the order is not important, it is a combination problem, and the answer is

7C5 = 21

Example $$\PageIndex{4}$$

How many line segments can be drawn by connecting any two of the six points that lie on the circumference of a circle?

Solution

Since the line that goes from point A to point B is same as the one that goes from B to A, this is a combination problem.

It is a combination of 6 objects taken 2 at a time. Therefore, the answer is

$6 \mathrm{C} 2=\frac{6 !}{4 ! 2 !}=15 \nonumber$

Example $$\PageIndex{5}$$

There are ten people at a party. If they all shake hands, how many hand-shakes are possible?

Solution

Note that between any two people there is only one hand shake. Therefore, we have

10C2 = 45 hand-shakes.

Example $$\PageIndex{6}$$

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?

Solution

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.
 B A

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 to be the horizontal blocks

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 10C5 = $$\frac{10!}{5! 5!}$$.

Example $$\PageIndex{7}$$

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

Solution

First we solve this problem using section 6.5 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

6C4 = 15.

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

6C2 = 15.

Further observe that by definition

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

and

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

Which implies 6C4 = 6C2.

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

Example $$\PageIndex{8}$$

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

Solution

We list 4 men and 4 women as follows:

$M_1M_2M_3M_4W_1W_2W_3W_4 \nonumber$

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:

 2-Man Committees 3-Woman Committees $\begin{array}{l} \mathrm{M}_{1} \mathrm{M}_{2} \\ \mathrm{M}_{1} \mathrm{M}_{3} \\ \mathrm{M}_{1} \mathrm{M}_{4} \\ \mathrm{M}_{2} \mathrm{M}_{3} \\ \mathrm{M}_{2} \mathrm{M}_{4} \\ \mathrm{M}_{3} \mathrm{M}_{4} \end{array} \nonumber$ $\begin{array}{l} \mathrm{W}_{1} \mathrm{W}_{2} \mathrm{W}_{3} \\ \mathrm{W}_{1} \mathrm{W}_{2} \mathrm{W}_{4} \\ \mathrm{W}_{1} \mathrm{W}_{3} \mathrm{W}_{4} \\ \mathrm{W}_{2} \mathrm{W}_{3} \mathrm{W}_{4} \end{array} \nonumber$

For every 2-man committee there are four 3-woman committees that can be chosen to make a 5-person committee. If we choose $$M_1M_2$$ as our 2-man committee, then we can choose any of $$W_1W_2W_3$$, $$W_1W_2W_4$$, $$W_1W_3W_4$$, or $$W_2W_3W_4$$ as our 3-woman committees. As a result, we get

$\boxed{M_1M_2}W_1W_2W_3, \boxed{M_1M_2} W_1W_2W_4, \boxed{M_1M_2} W_1W_3W_4, \boxed{M_1M_2} W_2W_3W_4 \nonumber$

Similarly, if we choose $$M_1M_3$$ as our 2-man committee, then, again, we can choose any of $$W_1W_2W_3$$, $$W_1W_2W_4$$, $$W_1W_3W_4$$, or $$W_2W_3W_4$$ as our 3-woman committees.

$\boxed{M_1M_3}W_1W_2W_3, \boxed{M_1M_3} W_1W_2W_4, \boxed{M_1M_3}W_1W_3W_4, \boxed{M_1M_3}W_2W_3W_4 \nonumber$

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 $$\PageIndex{9}$$

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

1. One student from each class?
2. All juniors?
3. Two freshmen and 2 seniors?
4. No freshmen?
5. At least three seniors?

Solution

a. Applying the multiplication axiom to the combinations involved, we get

( 4C1 ) ( 5C1 ) ( 5C1 ) ( 6C1 ) = 600

b. We are choosing all 4 members from the 5 juniors, and none from the others.

5C4 = 5

c. 4C2 $$\cdot$$ 6C2 = 90

d. Since we don't want any freshmen on the committee, we need to choose all members from the remaining 16. That is

16C4 = 1820

e. 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.

( 6C3 ) ( 14C1 ) + 6C4 = 295

Example $$\PageIndex{10}$$

How many five-letter word sequences consisting of 2 vowels and 3 consonants can be formed from the letters of the word INTRODUCE?

Solution

First we select a group of five letters consisting of 2 vowels and 3 consonants.
Since there are 4 vowels and 5 consonants, we have

( 4C2 ) ( 5C3 )

Since our next task is to make word sequences out of these letters, we multiply these by 5!.

( 4C2 ) ( 5C3 ) ( 5! ) = 7200.

Example $$\PageIndex{11}$$

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 card of another suit be drawn?

Solution

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

 Ways of selecting the first suit Ways of selecting 4 cards from this suit Ways of selecting the next suit Ways of selecting a card from that suit 4C1 13C4 3C1 13C1

( 4C1 ) ( 13C4 ) ( 3C1 ) ( 13C1 ) = 111,540.

## A STANDARD DECK OF 52 PLAYING CARDS

As in the previous example, many examples and homework problems in this book refer to a standard deck of 52 playing cards. Before we end this section, we take a minute to describe a standard deck of playing cards, as some readers may not be familiar with this.

A standard deck of 52 playing cards has 4 suits with 13 cards in each suit.

Each suit is associated with a color, either black (spades, clubs) or red (diamonds, hearts).

Each suit contains 13 denominations (or values) for cards:

an Ace (A), nine numbers 2, 3, 4, …., 10, and Jack(J), Queen (Q), King (K).

The Jack, Queen and King are called “face cards” because they have pictures on them. Therefore a standard deck has 12 face cards: (3 values J, Q, K ) x (4 suits ♦, ♥, ♠,  ♣ )

We can visualize the 52 cards by the following display

 Suit Color Values (Denominations) ♦ Diamonds Red A 2 3 4 5 6 7 8 9 10 J Q K ♥ Hearts Red A 2 3 4 5 6 7 8 9 10 J Q K ♠ Spades Black A 2 3 4 5 6 7 8 9 10 J Q K ♣ Clubs Black A 2 3 4 5 6 7 8 9 10 J Q K

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