First off, let me say that you can find the answer to this question in Sage using the nauty generator. If you're going to be a serious graph theory student, Sage could be very helpful.
count = 0
for g in graphs.nauty_geng("20 180:180"):
count = count+1
print count
The answer is 4613. But, this isn't easy to see without a computer program.
At this point, perhaps it would be good to start by thinking in terms of of the number of connected graphs with at most 10 edges. Then, all the graphs you are looking for will be unions of these. You should be able to figure out these smaller cases. If any are too hard for you, these are more likely to be in some table somewhere, so you can look them up.
Connected graphs of order n and k edges is:
n = 1, k = 0: 1
n = 2, k = 1: 1
n = 3, k = 2: 1
n = 3, k = 3: 1
n = 4, k = 3: 2
n = 4, k = 4: 2
n = 4, k = 5: 1
n = 4, k = 6: 1
n = 5, k = 4: 3
n = 5, k = 5: 5
n = 5, k = 6: 5
n = 5, k = 7: 4
n = 5, k = 8: 2
n = 5, k = 9: 1
n = 5, k = 10: 1
.
.
.
n = 10, k = 9: 106
n = 10, k = 10: 657
n = 11, k = 10: 235
I used Sage for the last 3, I admit. But, I do know that the Atlas of Graphs contains all of these except for the last one, on P7.
Notice that the sum of the number of edges in the original gragh $G$ and the complement graph $G'$ must be equal to the number of edges in a complete graph with the number of vertices being the number of vertices in $G$. This is because $G'$ will only contain edges not included in $G$.
Thus if the number of vertices in $G$ is $V$, then the total number of edges is $\frac{V(V-1)}{2}$. Since
$$
15+13 = 28 = \frac{56}{2} = \frac{8(8-1)}{2}
$$
Therefore there must be 8 vertices in the graph.
Best Answer
If you treat isomorphic graphs separately, then one way to think about this is as follows. If you start off with $N$ nodes and want $E$ edges, you could build a graph by considering an $E$-element subset of the set of all possible edges in the graph. The number of possible edges is $N^2$, since each node can be connected to each other node (including itself), so the number of ways you can pick $E$ of these edges is $\binom{N^2}{E}$. This is
$$ \frac{(N^2)!}{(E!) \cdot (N^2 - E)!}, $$
which is a staggeringly huge number.
Hope this helps!