Shatar: Imagine I had a complete graph and wanted to direct all the edges so that the number of incoming and outgoing edges to each vertex is the same.
JJ: You better have an odd number of vertices, or else it’s not going to happen.
Shatar: ok, Sure. But how many ways could I do it?
JJ: \((n-1)!\)
Shatar: Hey, no spoilers!

Bijection direction 1:

• Take a permutation of \(n-1\). place the numbers on the vertices of your graph, and put the number \(0\) on vertex \(v_0\).
• Draw an edge if \(i\to j\) is shorter walk \(\mod n\) than \(j\to i\)
• claim: this gets the indegree = outdegree constraint thing to happen

Now the other direction:

• Take a direction of the edges
• Let \(A\) be the out neighbors of \(v_0\).
• Give the number to \(v\in A\) based on how many other \(w\in A\) \(v\) points to
• do the same thing for the in-neighbors of \(v_0\)

So its \((n-1)!\)

gotta bounce type this better later?