Question

No of simple directed graphs are possible with 4 vertices and 3 edges

Answer 1

Let's solve this the same way we solve it for undirected graphs: For each pair of vertices (v,u) there are exactly four possibilities for the edges between them: either there is no edge between them, or just v->u, or just u->v, or both v->u and u->v. so the answer would be 4 in the power of the amount of pairs of vertices, that is 4^(n choose 2). Your idea was quite close, although you need to notice that the amount of ways to "choose 2 elements from a set of size N" is not 2N, but N^2. (which in our case gives the correct answer: (2^(n choose 2))^2 = 4^(n choose 2) )