TITLE: Give a combinatorial proof for a multiset identity
QUESTION [7 upvotes]: I'm asked to give a combinatorial proof of the following, $\binom{\binom n2}{2}$ = 3$\binom{n}{4}$ + n$\binom{n-1}{2}$.
I know $\binom{n}{k}$ = $\frac{n!}{k!(n-k)!}$ and $(\binom{n}{k}) = \binom{n+k-1}{k}$ but I'm at a loss as to what to do with the $\binom{\binom n2}{2}$
Can someone point me in the right direction as to how to proceed with writing a combinatorial proof for this identity?

REPLY [6 votes]: Consider a graph $G=(V,E)$ such that $|V|=n$ and $|E|={n \choose 2}$ (i.e. any two vertices are connected). Than the LHS of your identity is the number of unordered pairs of edges. Any such pair is either determined by four vertices or by three. In the first case we first choose 4 vertices and then connect them in any of 3 ways which gives us altogether $3{n \choose 4}$ combinations. In the second case we first select the vertex which belongs to both edges and later we select the other two vertices from remaining $n-1$ vertices which gives us $n{n-1 \choose 2}$ possibilities. Summing it up we get the RHS.