# The camera club has five members, and the mathematics club has eight. There is only one member common to both clubs. In how many ways could a committee of four people be formed with at least one member from each club?

1173

#### Explanation:

We have the camera club, which I'll dub P for photography, has 5 members. We also have the Mathematics club with 8 members. And in common with both of those, we have C - who is the one person common to both.

All this means that we have 4 unique members of P, 7 of M and 1 of C.

From this, we're going to pick 4 people and want at least 1 person from each group (P and M). There are two ways this can happen - either C gets picked (and so automatically we have representation from both groups), or C isn't picked and so we need 1 person each from P and M.

C is picked

If we have C be picked, there are now 11 members of P and M we can pick 3 people from:

C_(n,k)=((n),(k))=(n!)/((k!)(n-k)!) with $n = \text{population", k="picks}$

C_(11,3)=((11),(3))=(11!)/((3!)(8!))=165

If C is not picked

We need to pick one person from P, which is $\left(\begin{matrix}4 \\ 1\end{matrix}\right) = 4$, and one from M, which is $\left(\begin{matrix}7 \\ 1\end{matrix}\right) = 7$, and then 2 people from the remaining 9, which is $\left(\begin{matrix}9 \\ 2\end{matrix}\right) = 36$:

$4 \times 7 \times 36 = 1008$

Putting it together

$165 + 1008 = 1173$