Number of edges in a transformation graph

Question

Having a problem that came up with using an old exercise book(so i have no solution to check or if is valid).

for a simple non-directed graph $G$ of order $n$ and size $m>1,L(G)$ (another graph) is defined as:

a)Every vertex of $L(G)$ is an edge of $G\ (1-1$ function)

b)$2$ vertices of $L(G)$ are connected by an edge iff the equivalent edges in $G$ have a common vertex

Prove that the number of edges of $L(G)$ $$\sum_{u\in V(G)}\binom{d(u)}{2}$$(binomial coeficient).

My workprocess so far:By having a $1-1$ ratio of vertices $(L(G)$ - edges($G$) i can show that $n$ of $L(G)$ is equal to $\sum d(u)/2$. I need the concept especialy of b) explained to me.

Answer 1

So the degree of each vertex $e= \{x,y\}$ in $L(G)$ is $$d_L(e) = d(x)+d(y)-2$$

So by handshakeing lemma we have $${1\over 2}\sum _{\{x,y\}\in E}d_L(\{x,y\})= {1\over 2}\sum _{\{x,y\}\in E} (d(x)+d(y)-2)$$

Now since each $x$ appears in sum exactly $d(x)$ times we have:

$${1\over 2}\Big(\sum _{x\in V} d(x)^2\Big)-\varepsilon \;\;\;(*) $$

By handshakeing lemma for the starting graph we have $$\sum _{x\in V} d(x)=2\varepsilon $$

so we have $$ (*) = {1\over 2}\Big(\sum _{x\in V} d(x)^2-d(x) \Big) = \sum _{x\in V}{d(x)\choose 2} $$

Answer 2

Consider $v\in V(G)$.

If $d(v)\le1$, $v$ doesn't contribute any edge to $L(G)$ because we need at-least two edges incident on $v$ for it to be their common vertex.

Whereas if $d(v)>1,v$ has $d(v)$ distinct edges incident on it. Selecting any two of these edges gives us an edge in $L(G)$. Recall that two edges can be selected in $\binom{d(v)}2$ ways. We now sum this value over all the vertices of $G$.

Note that if $d(v)\le1,\binom{d(v)}2$ is assumed to be $0$.