I got the following question -
what is the number of 2-regular graphs containing an Euler cycle with n vertices
what I came up with so far -
as I understand we are looking for a circle (every vertex is of degree of 2)
so let's say we pick some random vertex - we have to pick his neighbors out of $n-1$ vertices - $$\left(\begin{array}{c}n-1\\ 2\end{array}\right)$$
and then we have to fill all the other vertices untill we reach the end of the cicrcle - $$(n-3)!$$ so the final answer would be $$\left(\begin{array}{c}n-1\\ 2\end{array}\right)\cdot (n-3)!$$
It works for $n=3$ and $n=4$ and I think it works for $n=5$
what do you think? what is wrong with my logic here? how I solve this kind of problems?
A 2-regular graph which has an Euler cycle with $n$ vertices is necessarily connected and therefore it is a cycle-graph with $n$ vertices. If we are counting the labeled graphs then the number should be for $n>2$ $$\frac{n!\;\mbox{(permutations)}}{2\cdot n\; \mbox{(orientations)(rotations)}}=\frac{(n-1)!}{2}$$ See OEIS's A001710.