How many directed graphs are possible using V vertices?

Question

As an example, if I have 2 vertices V1, V2. All the possible directed graphs ( there can be indegree 0 vertices as well) are 3 {{V1,V2}, {V1->V2}, {V1<-V2}}. But, when there are 3 vertices all the possible graphs are found to be very large. {{V1,V2,V3}, {V1->V2,V3}, {V1<-V2,V3}.....{V1->V2->V3} .... } . Like

1. graphs with edges 0
2. graphs with 1 edges and their directionality changed
3. graphs with 2 edges and their directionality changed
4. graphs with 3 edges and there directionality changed

My attempt so far: As mentioned above I simply analyze the questions, but it is found to be growing larger. I was able to found that for any N no. of vertices, all the possible edges ( undirected ) are N*(N-1) /2. It is like I have to analyze all the number of possible edges and their directionality.

But, which is not clear is the possible number of graphs is growing with a possible number of edges. I would appreciate any suggestions on how to obtain all the possible directed graphs from N number of vertices.

Answer 1

There are $\binom{V}{2}=V(V-1)/2$ distincy pairs of vertices. For any pair of vertices $\{u,v\}$, any one of the 3 options can hold:

1. Either there is no edge between $u$ and $v$
2. There is an edge directed from $u$ to $v$
3. There is an edge directed from $v$ to $u$

Hence total number of directed graphs on $V$ vertices is $3^{V(V-1)/2}$.

Answer 2

With $N$ vertices, there are $N_E = \binom{N}{2}$ possible edges. We can then get $\binom{N_E}{m}$ many undirected graphs with $m$ edges. Each edge has 2 possible directions giving $2^m\times\binom{N_E}{m}$ possible directed graphs with $m$ edges (because each undirected graph has $2^m$ possible orientations). Summing everything, you get $\sum_{m=0}^{N_E}\binom{N_E}{m}2^m = 3^{N_E} = 3^\binom{N}{2}$.