2008
Problem - 1607
Two subsets of the set $S=\{ a,b,c,d,e\}$ are to be chosen so that their union is $S$ and their intersection contains exactly two elements. In how many ways can this be done, assuming that the order in which the subsets are chosen does not matter?
Firstly, there are $C_5^2=10$ choices for the two common elements. It then leaves $3$ remaining elements to be distributed between two subsets. For each element there are two choices, hence the total count is $2^3=8$ possibilities. However, by the principle of symmetry (because the two sets can switch), there is a duplicate factor of $2$. This means the final answer is $$\frac{10\times 8}{2}=\boxed{40}$$