Verify that $\binom{n+1}{4} = \frac{\left(\substack{\binom{n}{2}\\{\displaystyle2}}\right)}{3}$ for $n \geq 4$ 
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.
 A: "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.
A: To verify:
$${n\choose2} = \frac{n(n-1)}{2}$$
$${\frac{n(n-1)}{2}\choose2} = \frac{\frac{n(n-1)}{2}.(\frac{n(n-1)}{2}-1)}{2}$$
$${\frac{n(n-1)}{2}\choose2} = \frac{n(n-1).(n^2-n-2)}{8} = \frac{(n+1)n(n-1)(n-2)}{8}\tag1 - RHS$$
$$ {(n+1)\choose4} = \frac{(n+1)n(n-1)(n-2)}{4!} = \frac{RHS}{3}$$
Hence proved.
A: Using 
\begin{align}
\binom{n}{2} = \frac{n(n-1)}{2}
\end{align}
then 
\begin{align}
\frac{1}{3} \, \binom{\binom{n}{2}}{2} &= \frac{1}{2} \binom{n}{2} \, \left(\frac{n}{2} - 1\right) \\
&= \frac{n(n-1)}{4!} \left( n(n-1)-2 \right) = \frac{(n+1)(n)(n-1)(n-2)}{4!} \\
&= \binom{n+1}{4}. 
\end{align}

As to the modified component of the question: The comments seem to be helpful
A: Algebraic Manipulation is actually not that bad. $\binom{n}{2}=n(n-1)/2$, so
$$\binom{\binom{n}{2}}{2}=\binom{n(n-1)/2}{2}=(n(n-1)/2)(n(n-1)/2-1)/6=n(n-1)(n^2-n-2)/24=n(n-1)(n+1)(n-2)/24=(n+1)!/4!(n-3)!=\binom{n+1}{4}$$
