**A little bit of background:**

In a small course a friend of mine is attending, everybody shall give a presentation about some topic. Now for each presentation, here can be two *assistants* which are other students that inform themselves about the topic and help moderating the discussion etc.

Of course, optimally, each participant gets two assistants for his presentation, and does assist in two others.

Surely, our combinatorics sense awoke and we wondered "How many possibilities are there of arranging studends, assistants and presentations together in the fair way?" ;)

**The model:**

An idea of modeling the situation was the following:

Let each student be represented by a vertex in a directed graph $G$. Then we draw an arc from $a$ to $b$ iff $a$ is assisting in $b$'s presentation.

Now the course situation in $G$ is fair if any vertex has two arcs going out and two coming in. In other words, $G$ is `2-2`

-regular. Thus, our problem comes down to

# The Question

How manys 2-2 regular directed graphs are there on $n$ vertices?

If we call this number $R_n$, we had tried a recursive approach. Introduce a new vertex to an existing regular graph, then it needs two arcs that have to be removed elsewhere. We can have $\binom n 2$ possibilities of vertices to take them away, and 2 vertices each, so we'd get

$$ R_{n+1} = 4 \binom n 2 R_n $$

Of course this is wrong, we generate degenerate possibilities with this approach that we need to exclude, and that's where we run into problems.

So, any ideas on counting these `2-2`

-regular graphs, please?

Enumerative Combinatorics, vol. 2, Cor. 5.5.11. Your problem introduces the extra requirement that $\mathrm{trace}(M)=0$, which makes things much more difficult, but conceivably the technique in the above reference can still be used. It looks to me that some variant of the ménage problem will be relevant. $\endgroup$ – Richard Stanley Nov 8 '12 at 19:51