TITLE: Is there a "counting groups/committees" proof for the identity $\binom{\binom{n}{2}}{2}=3\binom{n+1}{4}$?
QUESTION [8 upvotes]: 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?

REPLY [8 votes]: For the right hand side, add a special element $s$ to your $n$-element set; then choose $4$ elements from the extended set, and a partition of those $4$ into $2$ sets of size $2$ (the latter is possible in $3$ ways). If $s$ was not among the selected elements retain the two disjoint pairs; otherwise let the pairs be $\{s,x\}$ and $\{y,z\}$, and retain the sets $\{x,y\}$ and $\{x,z\}$. Every pair of pairs in the left hand side is counted once.