How many ways to choose the group to play tennis are there?

How many ways to choose the group to play tennis are there?

Question

There are $10$ people who are going to spend the afternoon in $2$ groups, one group to go to the theatre and the other to play tennis. In how many ways can the group for tennis be selected if there must be at least $4$ people in each group?

Solution

If there must be at least $4$ people in each group, then the number of people in the group for tennis must be $4$, $5$, or $6$.

There are

$$\begin{equation*} \binom{10}{k} = \frac{10!}{k!(10-k)!} \end{equation*}$$

ways of choosing

$k$

people from a group of

$10$

, and so there are

$$\begin{align*} \binom{10}{4} + \binom{10}{5} + \binom{10}{6} &= \frac{10!}{4!6!} + \frac{10!}{5!5!} + \frac{10!}{6!4!} \\ &= 2 \times \frac{10 \times 9 \times 8 \times 7}{4 \times 3 \times 2 \times 1} + \frac{10 \times 9 \times 8 \times 7 \times 6}{5 \times 4 \times 3 \times 2 \times 1} \\ &= 2 \times \frac{10 \times 3 \times 1 \times 7}{1 \times 1 \times 1 \times 1} + \frac{1 \times 3 \times 2 \times 7 \times 6}{1 \times 1 \times 1 \times 1 \times 1} \\ &= 420 + 252 \\ &= 672 \end{align*}$$

ways of choosing the groups.

You may disagree with this. If you think a group of $5$ going off to play tennis is a bit daft then of course you would get fewer combinations. $\binom{10}{4} + \binom{10}{6} = \frac{10!}{4!6!} + \frac{10!}{6!4!} = 2 \times \frac{10 \times 9 \times 8 \times 7}{4 \times 3 \times 2 \times 1} = 420$