In the diagram below, the four circles represent the four committees and member A is in two of them.

Adding members B, C and D in an obvious way gives the diagram below.

Now, each committee shares exactly one member with the committees next to it, but not with the one opposite. Adding members E and F gives the diagram below - showing that there are a total of 6 members.

In the diagram below, the four circles represent the four committees. The line represents a member.

As each member of the club is in two committees, each line should connect two circles.

As each pair of committees has exactly one member in common, each circle should be joined to each other circle by one line. This is shown in the diagram below, which contains 6 lines, so there are 6 members in the club.

Labelling the committees A, B, C and D can help us to count the members.

Committee A shares a member with each of the other committees, so contains 3 members.

Additionally, committee B shares a member with committees C and D - which is 2 more members.

Committees C and D also share a member.

Each member is only in two committees, so each of the members counted above must be a different person.

That gives a total of 3 + 2 + 1 = 6 members.

This information is also represented in the two-way table below.

A | B | C | D | |

A | - | |||

B | - | - | ||

C | - | - | - | |

D | - | - | - | - |

Each committee shares exactly one member with each other committee - so must contain 3 members.

3 $\times$ 4 = 12, so that makes 12 members altogether. But each member is in two committees - so each member has been counted twice.

So there are 12 $\div$ 2 = 6 members in the club.

You can find more short problems, arranged by curriculum topic, in our short problems collection.