Skill: Calculate combinations

Use a combination when you select objects and the order of selection does not matter. For example, choosing Alice, Ben, and Carla is the same group as choosing Carla, Alice, and Ben.

The number of ways to choose $r$ objects from $n$ distinct objects is

$$\binom{n}{r}=\frac{n!}{r!(n-r)!}.$$

The factors $(r!)$ remove the repeated counting caused by arranging the same selected objects in different orders.

Example: A club has $8$ members. How many different committees of $3$ members can be chosen?

Since a committee is a group, not an ordered list, use a combination:

$$\binom{8}{3}=\frac{8!}{3!(8-3)!}$$

Simplify the factorials:

$$\binom{8}{3} =\frac{8!}{3!5!} =\frac{8\cdot7\cdot6\cdot5!}{(3\cdot2\cdot1)5!}$$

Cancel $(5!)$:

$$\binom{8}{3} =\frac{8\cdot7\cdot6}{3\cdot2\cdot1} =\frac{336}{6} =56$$

Therefore, there are

$$\boxed{56}$$

different committees.

Remember that combinations have the symmetry property

$$\binom{n}{r}=\binom{n}{n-r}.$$

So, in this example, $\binom{8}{3}=\binom{8}{5}$. Choosing $3$ committee members is equivalent to choosing the $5$ members who are not on the committee.