So I need to count how many Eulerian cycles exist in the complete graph $K_{n,2}$. Two cycles are considered identical if:
1) The order of the edges is identical in both cycles. For example: The cycles $cabc, bcab, abca$ are identical.
Or
2) If we get the first cycle by reversing the order of the edges in the second cycle. For example: $abca$ and $acba$ are the same cycle.
Note: The order of the edges is what matters, not the vertices. Also, of course if $n$ is odd, there's no such cycle.
So I tried this in different ways and arrived at different solutions:
I can think of this as all the permutations of the side with $n$ vertices, removing all the reverse permutations, so that's $\frac {n!}{2}$.
Or, I can think of it as permutations in a cycle, which means $\frac {(n-1)!}{2}$.
Then I manually counted the Eulerian cycles in $K_{2,4}$ and I got 6, which works well with $(n-1)!$. So either I counted incorrectly, or $(n-1)!$ is the answer, but I don't understand why.
Any help would be appreciated.
Choose a starting vertex $v_1$ in the partite set $V$ of size $n$ and an edge $v_1x_1$ where $x_1$ is in the partite set $X$ of size $2$. We now delete the edge $v_1x_1$ from the graph and consider all Eulerian paths starting with $x_1$ and ending at $v_1$. This avoids counting identical cycles in the original graph.
Now everytime we choose an edge $x_iv_i$, we are forced to choose the next edge as $v_ix_j$ for $i\ne j$ because the graph is bipartite, and after we have done this there are no more edges remaining connected to $v_i$. This means that the only choice we have from the starting point is the order of vertices in $V\backslash \{v_1\}$ in our path (because we are forced to make $v_1$ last). This means the total number of orderings is $(n-1)!$.