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?

Question

Anthony Arnold · Accepted Answer

1173

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#