TITLE: A peculiar binomial coefficient identity
QUESTION [8 upvotes]: While inventing exercises for a discrete math text I'm writing I came up with this
$$
\binom{\binom{n}{2}}{2}=3\binom{n+1}{4}
$$
It's an easy result to prove, but it got me wondering

Is this pure coincidence, or is it a special case of some more general result of which I'm unaware?
No matter what the answer to (1) is, is there a combinatorial proof?

REPLY [2 votes]: You can count all of the pairs of pairs with $4$ distinct elements and all of the pairs of pairs with $3$ distinct elements and then add them together.
For the first sum get $\binom{n}{4}\binom{4}{2}\frac{1}{2}=3\binom{n}{4}$.  We divide by two because once we choose the first pair out of the $4$ elements we implicitly chose the other pair; we are double counting otherwise.
For the second sum we get $3\binom{n}{3}$ since we must choose $3$ elements and then one of them to be duplicated.
In total we have $$3 \left[ \binom{n}{4} + \binom{n}{3}\right] = 3\binom{n+1}{4}$$.
So that's $2/3$ combinatorial at least.