Is there a "counting groups/committees" proof for the identity $\binom{\binom{n}{2}}{2}=3\binom{n+1}{4}$? This is exercise number $57$ in Hugh Gordon's Discrete Probability. 

For $n \in \mathbb{N}$, show that
$$\binom{\binom{n}{2}}{2}=3\binom{n+1}{4}$$

My algebraic solution:
$$\binom{\binom{n}{2}}{2}=3\binom{n+1}{4}$$
$$\binom{\frac{n(n-1)}{2}}{2}=\frac{3n(n+1)(n-1)(n-2)}{4 \cdot 3 \cdot 2}$$
$$2\left(\frac{n(n-1)}{2}\right)\left(\frac{n(n-1)}{2}-1\right)=\frac{n(n+1)(n-1)(n-2)}{2}$$
$$2n(n-1)\frac{n^2-n-2}{2} = n(n+1)(n-1)(n-2)$$
$$n(n-1)(n-2)(n+1)=n(n+1)(n-1)(n-2)$$
This finishes the proof.

I feel like this is not what the point of the exercise was; it feels like an unclean, inelegant bashing with the factorial formula for binomial coefficients. Is there a nice counting argument to show the identity? Something involving committees perhaps?

$\binom{\binom n2}2$ counts pairs of (distinct) 2-element subsets of $n$-element set. Union of such pair is either 4-element set (and each 4-element set is counted 3 times: there are 3 ways to divide 4-set into 2 pairs) or 3-element set (and each 3-element set is also counted 3 times). That gives $3\binom n4+3\binom n3=3\binom{n+1}4$.
