How would I go about solving this?
I know that $K_{10}$ has $9+8+7+\dots+1=45$ edges.
So would it be something like $\binom {45}{43}$ because out of the 45 total edges, one must choose 43 for the graph?
Second attempt: At least for the second bullet point mentioned, it seems that there would be $\binom{10}{2}9\cdot7$ ways. There are two vertices to choose from ten. Then, for the first vertex, there are nine options for a connecting vertex, and for the second vertex, there are 7 options for a connecting vertex.}
Third attempt: I don't know what I was thinking ^. I can only find one non-isomorphic class for the second bullet point, and they are not isomorphic those in the first class.