A club of $n$ members is organized into four committees following two rules:

  • Each member belongs to exactly two committees, and

  • each pair of committees has exactly 1 member in common.

Find all possible values of $n$.


If each member must belong to two committees, he or she must be the member in common to those two committees. There are 6 pairs of committees, requiring 6 common memberships and therefore 6 members.

Any more members joining would give his or her two committees two common members.

Therefore, I would say only $n=6$.

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