How many graphs on n labeled vertices exist which have at least (n2 - 3n)/2…

Key idea: count graphs by the number of missing edges (the complement).

Let M = n(n-1)/2 be the total number of possible edges on n labeled vertices. Note that (n^2 - 3n)/2 = M - n, so a graph with at least (n^2 - 3n)/2 edges has at most n missing edges.

For each k with 0 ≤ k ≤ n, choose which k edges are missing from the M possible; choosing those missing edges uniquely determines the graph. There are C(M, k) choices for exactly k missing edges.

Therefore the total number of graphs with at least (n^2 - 3n)/2 edges is:

• Final answer: sum_{k=0}^{n} C(M, k) where M = n(n-1)/2.

This counts all graphs with at most n missing edges, which is equivalent to having at least (n^2 - 3n)/2 edges.