Question

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?

Answer 1

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="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 `((4),(1))=4`, and one from M, which is `((7),(1))=7`, and then 2 people from the remaining 9, which is `((9),(2))=36`:

`4xx7xx36=1008`

Putting it together

`165+1008=1173`