Problem: Prove that there's more than 6600 different classes of isomorphism of simple graphs with 8 vertices.
I really have no clue how to solve this.
I thought of counting how many edges can be in such graph (which is in an interval $\left<0, \binom{8}{2}\right>$.
Then I've tried to count the number of ways to put those edges in my graph. That should be equal to $\sum_{i=0}^{28}\binom{8}{2}^{i} = \sum_{i=0}^{28}28^{i}$
But that's where I've stopped because this counting lets me have more than one edge between two vertices, and I can't have that. Basically, my logic was wrong.
Any hints or comments are welcome!
Here’s a hint. The size of an isomorphism class is at most 8! and there are $2^\binom{8}2 = 2^{28}$ graphs with vertex set $\{1,\ldots,8\}$. So there are at least $2^{28}/8!$ isomorphism classes.