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.

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
