Let $a = \frac{n(n-1)}{2}, b = \frac{n^2 -3n}{2}$
Minimum no of edges has to be $\frac{n^2 -3n}{2} = b$.
Maximum no of edges in simple graph = $\frac{n(n-1)}{2} = a$.
So, no of graph with minimum $b$ edges :
$= C(a,b) + C(a,b+1) + C(a,b+2) + \dots +C(a,a)$
$= C(a,a-b) + C(a,a-(b+1)) + C(a,a-(b+2)) + \dots +C(a,0)$
$= C(a,n) + C(a,n-1) + C(a,n-2) + \dots +C(a,0))$$\;\left(\because a-b = n \right)$
$= C\left(\frac{n(n-1)}{2},n\right) + C\left(\frac{n(n-1)}{2}, n-1\right) \\+ C\left(\frac{n(n-1)}{2},n-2\right) + \dots +C\left(\frac{n(n-1)}{2},0\right)$
$ =\sum\limits_{k=0}^{n} {}^{\left(\frac{n^2-n}{2}\right)}C_k$
Option D..