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At the beginning of a paper by McKay and Robinson on enumerating eulerian circuits, the authors state that the number of regular tournaments containing a directed rooted tree $T$ on vertices $v_1,\dots,v_n$ with root $v_n$ coincides with the constant term in the generating function $$\prod_{1\le j<k\le n}(x_j^{-1}x_k+x_jx_k^{-1})\,\prod_{jk\in\textrm{ edges of }T}\frac{x_jx_k^{-1}}{x_j^{-1}x_k+x_jx_k^{-1}}\,.$$ Unlike most everything else in the paper, this statement is made without justification, which makes me think that it's either a well-known result or obvious, i.e., except to me.

Could someone provide a reference or a few words to justify this claim?

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The paper gives email addresses for each author. You could try contacting them –  Chris Godsil Mar 16 '12 at 19:37
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1 Answer

up vote 6 down vote accepted

The coefficient of $x_1^{i_1}\cdots x_n^{i_n}$ in $$\prod_{1\le j<k\le n}(x_j^{-1}x_k+x_jx_k^{-1})$$ is the number of tournaments on vertices 1, 2, ..., $n$ in which the outdegree minus the indegree of vertex $l$ is $i_l$ for each $l$. This is because each edge $\{j,k\}$ in the complete graph contributes either 1 to vertex $j$ and $-1$ to vertex $k$ or 1 to vertex $k$ and $-1$ to vertex $j$, corresponding to the factor $x_j^{-1}x_k+x_jx_k^{-1}$. So the constant term will count regular tournaments (every vertex has the same indegree as outdegree). If we require that the edge $\{j,k\}$ be directed from $j$ to $k$ then we replace the factor $x_j^{-1}x_k+x_jx_k^{-1}$ with $x_j x_k^{-1}$. (The fact that the edges form a directed rooted tree is not necessary.)

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Thank you, Ira. That helps very much. –  Russell May Mar 16 '12 at 21:26
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Thanks Ira, I couldn't have put it better myself ;). –  Brendan McKay Mar 17 '12 at 1:40
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