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Combination Problems

Using n!/r!(n - r)! to calculate permutations

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Evaluate Combinations Using Combination Notation

Let’s Think About It

Mauricio is waiting in line at the ice cream truck. There are nine different toppings available, and Mauricio has enough money to get two, but he can’t decide which two toppings to choose. How many different selections can Mauricio make?

In this concept, you will learn to use combination notation to evaluate combinations.

Guidance

Order is important for some groups of items but not important for others. Consider a list of the words: POTS, STOP, SPOT, and TOPS.

  •  For the spelling of each individual word, order is important. The words POTS, STOP, SPOT, and TOPS all use the same letters, but spell out very different words.
  •  For the list itself, order is not important. Whether the words are presented in one order –such as POTS, STOP, SPOT, TOPS, or another order, such as STOP, SPOT, TOPS, POTS, or a third order, such as TOPS, POTS, SPOT, and STOP -makes no difference. As long as the list includes all 4 words, the order of the 4 words doesn’t matter.

A combination is an arrangement of items in which order, or how the items are arranged, is not important. The collection of one order of the items is not functionally different than any other order.

When evaluating a combination, you can use a tree diagram. Use a tree diagram can be time consuming; combination notation is a much simpler option.

To use combination notation, you must first understand factorials. A factorial is a special number that represents the product of a set of values in descending order.

Let’s look at an example of a factorial evaluation.

Evaluate \begin{align*}5!\end{align*}.

You can say that this is the product of values starting with 5 in descending order.

\begin{align*}\begin{array}{rcl} 5! &=& 5 \times 4 \times 3 \times 2 \times 1 \\ &=&120 \\ \end{array}\end{align*}

The answer is 120.

You can use factorials and combination notation to evaluate combinations without using lists or tree diagrams.

Let’s take a look at how this works.

To represent the number of combinations there are for 6 items taken 4 at a time, use the notation \begin{align*}{\color{red}_6}C{\color{blue}_4}.\end{align*}

In general, combinations are written as:

\begin{align*}{\color{red}_n}C{\color{blue}_r}\end{align*}

where \begin{align*}{\color{red}n}\end{align*} items are taken \begin{align*}{\color{blue}r}\end{align*} at a time.

To compute \begin{align*} _nC_r\end{align*} use the formula:

\begin{align*}_nC_r = \frac{n!}{r!(n-r)!}\end{align*}

Let’s look at an example.

Find \begin{align*}_5C_2\end{align*}

First, set up the problem.

\begin{align*}\begin{array}{rcl} _nC_r = \frac{n!}{r!(n-r)!} \\ _5C_2 = \frac{5!}{2!(5-2)!} \\ \end{array}\end{align*}

Next, evaluate the expression.

\begin{align*}\begin{array}{rcl} _5C_2 &=& \frac{5!}{2!(5-2)!} \\ _5C_2 &=& \frac{5!}{2!(3)!} \\ _5C_2 &=& \frac{5 \times 4 \times 3 \times 2 \times 1}{2 \times 1(3 \times 2 \times 1)} \\ _5C_2 &=&\frac{5 \times 4}{2 \times 1} \\ _5C_2 &=&\frac{20}{2} \\ _5C_2 &=&10 \\ \end{array}\end{align*}

The answer is 10.

There are 10 combinations.

You can also use a graphing calculator to find combinations. For large numbers, especially, the calculator can save you time.

If you push the “m” button and on the top of the screen you will see PROB. #3 under the PROB menu is combinations.

Let’s try an example.

Find \begin{align*}_{25}C_{10}\end{align*} using your calculator. Notice the key press history to help you with the keystrokes.

The answer is 3,268,760.

Guided Practice

Write the following situation using combination notation. Then evaluate it.

Sixteen students went to the park. Four students could ride in four cars. How many different combinations of students could there be?

First, use combination notation to describe the problem.

Find \begin{align*}_{16}C_4\end{align*}

Next, evaluate the combination by simplifying first.

\begin{align*}\begin{array}{rcl} _nC_r &=& \frac{n!}{r!(n-r)!} \\ _{16}C_4 &=& \frac{16!}{4!(16-4)!} \\ _{16}C_4 &=& \frac{16!}{4!(12)!} \\ _{16}C_4 &=& \frac{16\times 15\times 14\times 13\times 12\times 11 \times 10\times 9 \times 8 \times 7\times 6\times 5\times 4\times 3\times 2\times 1}{4\times 3\times 2\times 1 \times (12\times 11 \times 10\times 9 \times 8 \times 7\times 6\times 5\times 4\times 3\times 2\times 1)} \\ _{16}C_4 &=& \frac{16\times 15\times 14\times 13}{4\times 3\times 2\times 1} \\ _{16}C_4 &=&\frac{43680}{24} \\ _{16}C_4 &=&1820 \\ \end{array}\end{align*}

The answer is 1,820.

There are 1820 different combinations.

Examples

Example 1

Find \begin{align*}_6C_3.\end{align*}

First, set up the problem.

\begin{align*}\begin{array}{rcl} _nC_r &=& \frac{n!}{r!(n-r)!} \\ _6C_3 &=& \frac{6!}{3!(6-3)!} \\ \end{array}\end{align*}

Next, evaluate the expression.

\begin{align*}\begin{array}{rcl} _6C_3 &=& \frac{6!}{3!(6-3)!} \\ _6C_3 &=& \frac{6!}{3!(3)!} \\ _6C_3 &=& \frac{6 \times 5 \times 4 \times 3 \times 2 \times 1}{3\times 2\times 1\times (3\times 2\times 1)}\\ _6C_3 &=& \frac{6\times 5 \times4}{3\times 2\times 1} \\ _6C_3 &=& \frac{120}{6} \\ _6C_3 &=& 20 \\ \end{array}\end{align*}

The answer is 20.

There are 20 combinations.

Example 2

Find \begin{align*}_9C_3.\end{align*}

First, set up the problem.

\begin{align*}\begin{array}{rcl} _nC_r &=& \frac{n!}{r!(n-r)!} \\ _9C_3 &=&\frac{9!}{3!(9-3)!} \\ \end{array}\end{align*}

Next, evaluate the expression.

\begin{align*}\begin{array}{rcl} _9C_3 &=& \frac{9!}{3!(9-3)!} \\ _9C_3 &=& \frac{9!}{3!(6)!} \\ _9C_3 &=& \frac{9\times 8\times 7\times 6\times 5 \times 4 \times 3 \times 2 \times 1}{3 \times2 \times 1 \times (6\times 5\times 4\times 3 \times 2 \times 1)} \\ _9C_3 &=&\frac{9 \times 8 \times 7}{3 \times 2 \times 1} \\ _9C_3 &=&\frac{504}{6} \\ _9C_3 &=&84 \end{array}\end{align*}

The answer is 84.

There are 84 combinations.

Example 3

Find \begin{align*}_5C_4\end{align*}.

First, set up the problem.

\begin{align*}\begin{array}{rcl} _nC_r&=& \frac{n!}{r!(n-r)!} \\ _5C_4 &=&\frac{5!}{4!(5-4)!} \\ \end{array}\end{align*}

Next, evaluate the expression.

\begin{align*}\begin{array}{rcl} _5C_4 &=&\frac{5!}{4!(5-4)!} \\ _5C_4&=& \frac{5!}{4!(1)!} \\ _5C_4 &=& \frac{5 \times 4 \times 3 \times 2 \times 1} {4\times 3 \times2 \times 1 \times (1)} \\ _5C_4 &=&\frac{5}{1}\\ _5C_4 &=&5\\ \end{array}\end{align*}

The answer is 5.

There are 5 combinations.

Follow Up

Remember Mauricio and his toppings?

Mauricio has to choose two ice cream toppings from nine different toppings available. How many different combinations can he choose?

First, write the combination notation to describe the situation.

\begin{align*}_9C_2\end{align*}

Next, set up the problem.

\begin{align*}\begin{array}{rcl} _nC_r &=& \frac{n!}{r!(n-r)!} \\ _9C_2 &=& \frac{9!}{2!(9-2)!} \\ \end{array}\end{align*}

Then, evaluate the expression.

\begin{align*}\begin{array}{rcl} _9C_2 &=& \frac{9!}{2!(9-2)!} \\ _9C_2 &=& \frac{9!}{2!(7)!} \\ _9C_2 &=& \frac{9\times 8\times 7\times 6\times 5 \times 4 \times 3 \times 2 \times 1}{2 \times 1 \times (7 \times 6\times 5\times 4\times 3 \times 2 \times 1)} \\ _9C_2 &=&\frac{9 \times 8 }{2 \times 1} \\ _9C_2 &=&\frac{72}{2}\\ _9C_2 &=&36\\ \end{array}\end{align*}

The answer is 36.

There are 36 different combinations of two toppings.

Video Review

https://www.youtube.com/watch?v=SGn1913lOYM

Explore More

Evaluate each combination.

1. Find \begin{align*}_5C_2\end{align*}

2. Find \begin{align*}_6C_5\end{align*}

3. Find \begin{align*}_7C_2\end{align*}

4. Find \begin{align*}_7C_3\end{align*}

5. Find \begin{align*}_8C_2\end{align*} 

6. Find \begin{align*}_6C_4\end{align*}

7. Find \begin{align*}_9C_2\end{align*}

8. Find \begin{align*}_9C_4\end{align*}

9. Find \begin{align*} _8C_3\end{align*}

10. Find \begin{align*}_4C_4\end{align*}

Use the formula to figure out the different combinations.

11. How many different color pairs are there among red, orange, yellow, green, and blue?

12. How many different sets of 3 colors are there among red, orange, yellow, green, and blue?

13. How many different color pairs are there among red, orange, yellow, green, blue, and purple?

14. How many different sets of 3 colors are there among red, orange, yellow, green, blue, and purple?

15. How many different sets of 3 colors are there among red, orange, yellow, green, blue, purple, and white?

16. Ten tennis players are on the Davis Cup Team. Only two players can play in the doubles finals. How many different doubles teams could play in the finals?

Vocabulary

combination

combination

Combinations are distinct arrangements of a specified number of objects without regard to order of selection from a specified set.
combination notation

combination notation

Combination notation has the forms nCr and c(n, r) where n is the number of different units to choose from and r is the number of units in each group.
combinatorics

combinatorics

Combinatorics is the study of permutations and combinations.
factorial

factorial

The factorial of a whole number n is the product of the positive integers from 1 to n. The symbol "!" denotes factorial. n! = 1 \cdot 2 \cdot 3 \cdot 4...\cdot (n-1) \cdot n .
n value

n value

When calculating permutations with the TI calculator, the n value is the number of objects from which you are choosing.
permutation notation

permutation notation

Permutation notation is the form  nPr  or P(n, r), and indicates the number of ways that n objects can be ordered into groups of r items each.
TI-84

TI-84

The TI-84 calculator is a graphing calculator produced by Texas Instruments and is considered an “industry standard” for more advanced calculations.

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