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# 12.15: Combination Problems

Difficulty Level: Basic Created by: CK-12
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Remember the decorations in the last Concept? Take a look once again.

The decorating committee is getting the stage ready for the Talent Show. There was a bunch of different decorating supplies ordered, and the students on the committee are working on figuring out the best way to decorate the stage.

They have four different colors of streamers to use to decorate.

Red

Blue

Green

Yellow

“I think four is too many colors. How about if we choose three of the four colors to decorate with?” Keith asks the group.

“I like that idea,” Sara chimes in. “How many ways can we decorate the stage if we do that?”

The group begins to figure this out on a piece of paper.

Combinations are arrangements where order does not make a difference. The decorating committee is selected three colors from the possible four options. Therefore, the order of the colors doesn’t matter.

Combinations are the way to solve this problem. Look at the information in this Concept to learn how to figure out the possible combinations.

### Guidance

Once you figure out if you are going to be using permutations or combinations, it is necessary to count the combinations.

There are several different ways to count combinations. When counting, try to keep the following in mind:

• Go one by one through the items. Don’t stop your list until you’ve covered every possible link of one item to all other items.
• Keep in mind that order doesn’t matter. For combinations, there no difference between $AB$ and $BA$ . So if both $AB$ and $BA$ are on your list, cross one of the choices off your list.
• Check your list for repeats. If you accidentally listed a combination more than once, cross the extra listings off your list.

James needs to choose a 2-color combination for his intramural team t-shirts. How many different 2-color combinations can James make out of red, blue, and yellow?

One way to find the number of combinations is to make a tree diagram. Here, if red is chosen as one color, that leaves only blue and yellow for the second color.

The diagram shows all 6 permutations of the 3 colors. But wait–since we are counting COMBINATIONS here order doesn’t matter.

So in this tree diagram we will cross out all outcomes that are repeats. For example, the first red-blue is no different from blue-red, so we’ll cross out blue-red.

In all, there are 3 combinations that are not repeats.

This method of making a tree diagram and crossing out repeats is reliable, but it is not the only way to find combinations.

Let's look at a situation like this.

James has added a fourth color, green, to choose from in selecting a 2-color combination for his intramural team. How many different 2-color combinations can James make out of red, blue, yellow, and green?

Step 1 : Write the choices. Match the first choice, red, with the second, blue. Add the combination, red-blue, to your list. Match the other choices in turn. Add the combinations to your list.

Step 2 : Now move to the second choice, blue. Match blue up with every possible partner. Add the combinations to your list.

Step 3 : Now move to the third choice, yellow. There is only one combination left to match it with. Add the combination to your list.

Your list is now complete. There are 6 combinations.

Sometimes, you won’t want to use all of the possible options in the combination. Think about it as if you have 16 flavors of ice cream, but you only want to use three flavors at a time. This is an example where there are 16 flavors to work with, but you can only use three at a time. With an example like this one, you are looking for combinations of object where only a certain number of them are used in any one combination.

This happens a lot with teams.

Here is a situation with teams.

How many different 2-player soccer teams can Jean, Dean, Francine, Lurleen, and Doreen form?

$&\underline{\text{Combination}} \qquad \qquad \qquad \qquad \qquad \qquad \quad \ \underline{\text{List}}\\&\text{Jean, Dean, Francine, Lurleen, Doreen} \qquad \text{Jean-Dean}\\&\text{Jean, Dean, Francine, Lurleen, Doreen} \qquad \text{Jean-Francine}\\&\text{Jean, Dean, Francine, Lurleen, Doreen} \qquad \text{Jean-Lurleen}\\&\text{Jean, Dean, Francine, Lurleen, Doreen} \qquad \text{Jean-Doreen}$

Step 2 : You’ve covered all combinations that begin with Jean. Now go through all combinations that begin with Dean, Francine, and Lurleen.

$&\underline{\text{Combination}} \qquad \qquad \qquad \qquad \qquad \qquad \quad \ \underline{\text{List}}\\& \text{Jean, Dean, Francine, Lurleen, Doreen} \qquad \text{Jean-Dean}\\& \text{Jean, Dean, Francine, Lurleen, Doreen} \qquad \text{Jean-Francine}\\& \text{Jean, Dean, Francine, Lurleen, Doreen} \qquad \text{Jean-Lurleen}\\& \text{Jean, Dean, Francine, Lurleen, Doreen} \qquad \text{Jean-Doreen}\\& \text{Jean, Dean, Francine, Lurleen, Doreen} \qquad \text{Dean-Francine}\\& \text{Jean, Dean, Francine, Lurleen, Doreen} \qquad \text{Dean-Lurleen}\\& \text{Jean, Dean, Francine, Lurleen, Doreen} \qquad \text{Dean-Doreen}\\& \text{Jean, Dean, Francine, Lurleen, Doreen} \qquad \text{Francine-Lurleen}\\& \text{Jean, Dean, Francine, Lurleen, Doreen} \qquad \text{Francine-Doreen}\\& \text{Jean, Dean, Francine, Lurleen, Doreen} \qquad \text{Lurleen-Doreen}$

Your list is now complete. There are 10 combinations.

Look at this situation.

How many different 3-player soccer teams can Jean, Dean, Francine, Lurleen, and Doreen form?

Use the process above to go through all of the combinations.

$&\underline{\text{Combination}} \qquad \qquad \qquad \qquad \qquad \qquad \quad \ \underline{\text{List}}\\& \text{Jean, Dean, Francine, Lurleen, Doreen} \qquad \text{Jean-Dean-Francine}\\& \text{Jean, Dean, Francine, Lurleen, Doreen} \qquad \text{Jean-Dean-Lurleen}\\& \text{Jean, Dean, Francine, Lurleen, Doreen} \qquad \text{Jean-Dean-Doreen}\\& \text{Jean, Dean, Francine, Lurleen, Doreen} \qquad \text{Jean-Francine-Lurleen}\\& \text{Jean, Dean, Francine, Lurleen, Doreen} \qquad \text{Jean-Francine-Doreen}\\& \text{Jean, Dean, Francine, Lurleen, Doreen} \qquad \text{Jean-Lurleen-Doreen}\\& \text{Jean, Dean, Francine, Lurleen, Doreen} \qquad \text{Dean, Francine-Lurleen}\\& \text{Jean, Dean, Francine, Lurleen, Doreen} \qquad \text{Dean-Francine-Doreen}\\& \text{Jean, Dean, Francine, Lurleen, Doreen} \qquad \text{Dean-Lurleen-Doreen}\\& \text{Jean, Dean, Francine, Lurleen, Doreen} \qquad \text{Francine-Lurleen-Doreen}$

Your list is now complete. There are 10 combinations.

We can use a formula to help us to calculate combinations. This is very similar to the work that you did in the last section with factorials and permutations.

Suppose you have 5 marbles in a bag–red, blue, yellow, green, and white. You want to know how many combinations there are if you take 3 marbles out of the bag all at the same time. In combination notation you write this as:

${_5}C_3 \Longleftarrow 5 \ \text{items taken 3 at a time}$

In general, combinations are written as:

${_n}C_r \Longleftarrow n \ \text{items taken} \ r \ \text{at a time}$

To compute ${_n}C_r$ use the formula:

${_n}C_r = \frac{n!}{r!(n - r)!}$

This may seem a bit confusing, but it isn’t. Notice that the factorial symbol is used with the number of object $(n)$ and the number taken at any one time $(r)$ . This helps us to understand which value goes where in the formula.

Now let’s look at applying the formula to the example.

For ${_5}C_3$ :

${_5}C_3 = \frac{5!}{3!(5 - 3)!} = \frac{5!}{3! 2!}$

Simplify.

${_5}C_3 = \frac{5 (4)(3)(2)(1)}{(3 \cdot 2 \cdot 1)(2 \cdot 1)} = \frac{120}{12} = 10$

There are 10 possible combinations.

Here is another one.

Find ${_6}C_2$

Step 1 : Understand what ${_6}C_2$ means.

${_6}C_2 \Longleftarrow 6 \ \text{items taken 2 at a time}$

Step 2 : Set up the problem.

${_6}C_2 = \frac{6!}{2!(6 -2)!}$

Step 3 : Fill in the numbers and simplify.

${_6}C_2 = \frac{6(5)(4)(3)(2)(1)}{(2 \cdot 1)(4 \cdot 3 \cdot 2 \cdot 1)} = \frac{720}{48} = 15$

There are 15 possible combinations.

Find the number of combinations in each example.

#### Example A

${_5}C_2$

Solution: $30$

#### Example B

${_4}C_3$

Solution: $1$

#### Example C

${_6}C_4$

Solution: $15$

Here is the original problem once again.

The decorating committee is getting the stage ready for the Talent Show. There was a bunch of different decorating supplies ordered, and the students on the committee are working on figuring out the best way to decorate the stage.

They have four different colors of streamers to use to decorate.

Red

Blue

Green

Yellow

“I think four is too many colors. How about if we choose three of the four colors to decorate with?” Keith asks the group.

“I like that idea,” Sara chimes in. “How many ways can we decorate the stage if we do that?”

The group begins to figure this out on a piece of paper.

Combinations are arrangements where order does not make a difference. The decorating committee is selected three colors from the possible four options. Therefore, the order of the colors doesn’t matter.

We can use combination notation to figure out this problem.

${_4}C_3 = \frac{4!}{3!(4 - 3)!} = \frac{4(3)(2)(1)}{(3 \cdot 2 \cdot 1)(1)}= \frac{24}{6} = 4$

There are four possible ways to decorate the stage.

Now that the students have this information, they can look at their color choices and vote on which combination they like best.

### Vocabulary

Here are the vocabulary words used in this Concept.

Combination
an arrangement of objects or events where order does not matter.
Permutations
an arrangement of objects or events where the order does matter.

### Guided Practice

Here is one for you to try on your own.

${_5} C_4$

${_5}C_4 = \frac{5!}{4!(5 - 4)!} = \frac{5(4)(3)(2)(1)}{(4 \times 3 \times 2 \times 1)1}= \frac{120}{24} = 5$

### Video Review

Here is a video for review.

### Practice

Directions: Evaluate each factorial.

1. 5!

2. 4!

3. 3!

4. 8!

5. 9!

6. 6!

Directions: Evaluate each combination using combination notation.

7. ${_7} C_2$

8. ${_7} C_6$

9. ${_8} C_4$

10. ${_9} C_6$

11. ${_8} C_3$

12. ${_{10}}C_7$

13. ${_{12}}C_9$

14. ${_{11}}C_9$

15. ${_{16}}C_{14}$

Basic

Dec 21, 2012

Jun 13, 2014