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?
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. \[\binom{10}{4} + \binom{10}{6} = \frac{10!}{4!6!} + \frac{10!}{6!4!} = 2 \times \frac{10 \times 9 \times 8 \times 7}{4 \times 3 \times 2 \times 1} = 420\]
