TITLE: Verify that $\binom{n+1}{4} = \frac{\left(\substack{\binom{n}{2}\\{\displaystyle2}}\right)}{3}$ for $n \geq 4$
QUESTION [6 upvotes]: Verify that for $n \geq 4$
$$\dbinom{n+1}{4} = \frac{\left(\substack{\binom{n}{2}\\{\displaystyle2}}\right)}{3}$$
Now present a combinatoric argument for the above.

First, by verify does it mean check for some n > 3? if so, then n = 4 gives both sides value 5. I have tried expanding both sides and It just gets messy, but i'm sure that would be attempting to prove it?
I cant quite move ahead with this one, I have said that $\left(\substack{\binom{n}{2}\\{\displaystyle2}}\right)$ is the number of ways of choosing 2 pairs of objects from n objects. my reasoning for this is that $\dbinom{n}{2}$ is the number of ways of choosing 2 objects from n and hence  $\left(\substack{\binom{n}{2}\\{\displaystyle2}}\right)$ is the number of ways of choosing 2 of these ways... What i am struggling to do is understand how the three comes into it.

REPLY [5 votes]: "Verify" would mean verify algebraically (using formulas such as $\binom{n+1}{4}=\frac{(n+1)n(n-1)(n-2)}{24}$. It's not that hard if you just consider the factorizations of each side. However, there is a nice combinatorial argument. $\left(\substack{{\binom{n}{2}}\\{\displaystyle 2}}\right)$ is the number of ways to pick two pairs of elements from a set of $n$; we can combine these pairs, and if an element is common to both pairs, then treat one of the instances of the element as a new $(n+1)$-st element. This counts the number of ways to pick 4 elements from a set of $n+1$. However, note that all ways to pick 4 are counted exactly 3 times, so you must divide by 3 to get the equality that is desired.