A simple graph with n vertices is connected if if has more than $\frac{(n-1)(n-2)}{2}$ edges.
The n vertex graph with the maximal number of edges that is still disconnected is a Kn−1
a complete graph Kn−1 with n−1 vertices has $\binom{n-1}{2}$edges, so $\frac{(n-1)(n-2)}{2}$ edges.
Adding any possible edge must connect the graph, so the minimum number of edges needed to guarantee connectivity for an n vertex graph is $\frac{(n-1)(n-2)}{2}$+1
Hence,Option(B)More than $\frac{(n-1)(n-2)}{2}$.