Finding the number of k - graphs on set of vertexes I am solving problem that i have quite hard time with.

How many non - isomorphic  10-regular graphs can we find on a 
   12 vertices?

I have only dealt with question about 2-regular graphs which is easy to find out considering we can create a combination of cirle and path.
What is the best way to solve a problem such as this? What facts should i consider?
 A: Suppose $G$ is a $10$-regular graph on $12$ vertices (we shall see that such a graph exists). Pick a vertex $v$. As $G$ is $10$-regular, $v$ has $10$ neighbours. So observe that there is exactly one vertex, say $w$, which is not $v$ and is not in the neighbourhood of $v$. Since $w$ also has $10$ neighbours, we see that the neighbourhood of $w$ is the same as the neighbourhood of $v$.
Now suppose we remove the vertices $v$ and $w$ from the graph. There are $10$ remaining vertices. Each of these vertices was adjacent to both $v$ and $w$ in $G$ and each had $10$ neighbours. So, after removing $v$ and $w$, the remaining graph is an $8$-regular graph on $10$ vertices.
Continue in this manner, we can remove a further to vertices to obtain a $6$-regular graph on $8$ vertices, then a $4$-regular graph on $6$ vertices, and finally a $2$-regular graph on $4$ vertices.
Therefore, the problem reduces to finding the number of non-isomorphic $2$-regular graphs on $4$ vertices. From your comment, the rest should be straightforward. Having found a $2$-regular graph on $4$ vertices, you can run this process backwards to obtain a $10$-regular graph on $12$ vertices.
You may also like trying to generalise this result to $2k$-regular graphs on $2(k+1)$ vertices.
