Note that the sum of the degrees of the graph is $23 \cdot 2=46$, so at least one vertex has degree 5 or more. We casework on the maximal degree $n$. Case 1: $n \geq 7$, then none of the $n$ neighbors can have an edge between each other, for $\binom{n}{2}$ edges unusable, and the vertex with maximal degree cannot connect to the $9-n$ other vertices. Then we have $\binom{n}{2}+9-n>\binom{10}{2}-23=22$ when $n \geq 7$, so there cannot be any graph in this case. Case 2: $n=6$. WLOG suppose that 1 is connected to $2,3,4,5,6,7$, then none of $2,3,4,5,6,7$ can connect to each other. Case 2.1: There is at least one edge between $8,9,10$, then each of $2,3,4,5,6,7$ can connect to at most two of $8,9,10$, for at most $6 \cdot 2+\binom{3}{2}=15$ additional edges. Along with the 6 original edges, it is not enough to each 23 edges. Case 2.2: There are no edges between $8,9,10$, then this graph is a bipartite graph between $1,8,9,10$ and $2,3,4,5,6,7$. There can be at most $4 \cdot 6=24$ edges in this graph, so exactly one edge is removed from this graph. There are $\binom{10}{4} \cdot 24=5040$ possible graphs in this case. Case 3: $n=5$. WLOG suppose that 1 is connected to $2,3,4,5,6$, then none of $2,3,4,5,6$ can connect to each other. Case 3.1: There is at least one edge between $7,8,9,10$. Then each of $2,3,4,5,6$ can connect to at most three of $7,8,9,10$, for $5 \cdot 3=15$ edges. In this case at least three of $7,8,9,10$ must not be connected to each other, so there can be at most three edges, for $5+15+3=23$ edges at most. However, this requires the three disconnected vertices of $7,8,9,10$ to be connected to all of $2,3,4,5,6$ and the other vertex of $7,8,9,10$, causing them to have degree 6. We can therefore ignore this case. (The case where $2,3,4,5,6$ can connect to two or less of $7,8,9,10$ can be easily ruled out.) Case 3.2: There are no edges between $7,8,9,10$, then this graph is a bipartite graph between $1,7,8,9,10$ and $2,3,4,5,6$. This is a $K_{5,5}$ with two edges removed, which accounts for $\binom{10}{5} / 2 \cdot\binom{25}{2}=126 \cdot 300=37800$ graphs. It is not difficult to see that Case 2.2 and Case 3.2 are disjoint (by considering max degree), so there are $5040+37800=42840$ graphs in total.