Combinations

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 [latex]r[/latex] objects from a set of [latex]n[/latex] objects where the order does not matter can be written as [latex]C\left(n,r\right)[/latex]. Just as with permutations, [latex]\text{C}\left(n,r\right)[/latex] can also be written as [latex]{}_{n}{C}_{r}[/latex]. In this case, the general formula is as follows.

[latex]\text{C}\left(n,r\right)=\dfrac{n!}{r!\left(n-r\right)!}[/latex]

formula for combinations of [latex]n[/latex] distinct objects

Given [latex]n[/latex] distinct objects, the number of ways to select [latex]r[/latex] objects where the order does not matter from the set is

[latex]\text{C}\left(n,r\right) = {}_{n}{C}_{r} =\dfrac{n!}{r!\left(n-r\right)!}[/latex]

An earlier problem considered choosing [latex]3[/latex] of [latex]4[/latex] possible paintings to hang on a wall. We found that there were [latex]24[/latex] ways to select [latex]3[/latex] of the [latex]4[/latex] paintings in order. But, what if we did not care about the order? We would expect a smaller number because selecting paintings [latex]1, 2, 3[/latex] would be the same as selecting paintings [latex]2, 3, 1[/latex].
[latex]\\[/latex]
To find the number of ways to select [latex]3[/latex] of the [latex]4[/latex] paintings, disregarding the order of the paintings, divide the number of permutations by the number of ways to order [latex]3[/latex] paintings.
[latex]\\[/latex]
There are [latex]3!=3\cdot 2\cdot 1=6[/latex] ways to order [latex]3[/latex] paintings.Thus, there are [latex]\frac{24}{6} = 4[/latex] ways to select [latex]3[/latex] of the [latex]4[/latex] paintings.
[latex]\\[/latex]
Using the formula:

[latex]\text{C}\left(4,3\right) = {}_{4}{C}_{3} =\dfrac{4!}{3!\left(4-3\right)!} = \dfrac{4 \cdot 3 \cdot 2 \cdot 1}{(3 \cdot 2 \cdot 1)(1)} = \dfrac{24}{6} = 4[/latex] ways.

Note the similarity and difference between the formulas for permutations and combinations:

• Permutations (order matters), [latex]P(n, r)=\dfrac{n!}{(n-r)!}[/latex]
• Combinations (order does not matter), [latex]C(n, r)=\dfrac{n!}{r!(n-r)!}[/latex]

The formula for combinations is the formula for permutations with the number of ways to order [latex]r[/latex] objects divided away from the result.

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

1. How many ways can you select your side dishes?
2. How many ways can you select [latex]3[/latex] side dishes?

Show Solution

1. We want to choose [latex]2[/latex] side dishes from [latex]5[/latex] options.
   [latex]\text{C}\left(5,2\right)=\dfrac{5!}{2!\left(5 - 2\right)!}=10[/latex]
2. We want to choose [latex]3[/latex] side dishes from 5 options.
   [latex]\text{C}\left(5,3\right)=\dfrac{5!}{3!\left(5 - 3\right)!}=10[/latex]

Using Calculator:
[latex]\\[/latex]
We can also use a graphing calculator to find combinations. Enter [latex]5[/latex], then press [latex]{}_{n}{C}_{r}[/latex], enter [latex]3[/latex], and then press the equal sign. The [latex]{}_{n}{C}_{r}[/latex], function may be located under the MATH menu with probability commands.