Is there a "counting groups/committees" proof for the identity $\binom{\binom{n}{2}}{2}=3\binom{n+1}{4}$?

Question

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?

Answer 1

$\binom{\binom n2}2$ 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 $3\binom n4+3\binom n3=3\binom{n+1}4$.

Answer 2

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.