Proof of simple graph using pigeonhole theorem Statement :
Any simple graph G with 
$$
 {\huge\Sigma}_{v\in V} {d(v)\choose{2}} > {{n}\choose{2}}
$$ 
contains a quadrilateral.

Proof: Denote by p2 the number of length two in G, and by p2(v) the number of such paths whose central vertex is v. 
Clearly $$p_2={d(v)\choose{2}}.$$
As each path of length two has a unique central vertex,$$p_2={\Sigma}_{v\in V}\:p_2(v)=\Sigma_{v\in V}{d(v)\choose{2}}.$$
On the other hand, each such path also has a unique pair of ends.
Therefore the set of all paths of length two can be partitioned into$${n}\choose{2}$$subsets according to their ends. The hypothesis $$\Sigma_{v\in V}{{d(v)}\choose{2}}>{{n}\choose{2}}$$now implies, by
virtue of the Pigeonhole Principle, that one of these subsets contains two or
more paths; that is, there exist two paths of length two with the same pair of
ends. 
The union of these paths is a quadrilateral. 
I am not able to understand how set of all paths of length two can be partitioned into ${{n}\choose{2}}$ subsets. If we're choosing any 2 vertices, how can it be insured that they have length 2 ?
If you decide to downvote the question, please provide a reason so that I can rectify the mistakes.
 A: The proof is assuming that such a graph exists with no quadrilateral, and deducing a contradiction. If a graph has no quadrilateral, it has at most $\binom n2$ paths of length two, since for each of the $\binom n2$ pairs of vertices $u,v$ there is at most one path of length two between $u$ and $v$ - if there were two such paths $uxv$ and $uyv$ then $uxvy$ would be a quadrilateral.
I think the confusion may be that it is "at most $\binom n2$", not "exactly $\binom n2$". There could be some pairs of vertices without any paths of length two, but that's not a problem.
A: Each path $u, uv, v, vw, w$ of length $2$ has two endpoints: $u$ and $w$. We partition the set of paths into $\binom n2$ subsets by the set of endpoints $\{u,w\}$. 
The subset of paths corresponding to $\{u,w\}$ is the set of all paths of length $2$ between $u$ and $w$, and may be empty (if there are no such paths), have size $1$ (if there is exactly one such path), or have size $2$ or more (if there are multiple such paths).
However, since the total number of paths is greater than $\binom n2$, and we divide the set of paths into $\binom n2$ subsets, there must be a subset corresponding to some pair $\{u,w\}$ which includes at least $2$ paths:
$$u, uv_1, v_1, v_1w, w\qquad \text{and} \qquad u, uv_2, v_2, v_2w, w.$$
Then the vertices $u, v_1, w, v_2$ span a quadrilateral.
