Combinatorial argument in Alon Tarsi Theorem In their proof of the main result of this paper: http://www.tau.ac.il/~nogaa/PDFS/chrom3.pdf
Alon and Tarsi provide a brief combinatorial argument to justify Lemma 2.2. That the co-efficient of the given monomial is the difference of the number of even and odd orientations of a graph G, where each orientation is contricted to keep each vertex $v_i$ with outdegree $d_i$. 
Why is this co-efficient as such? In my head the only way to reason like this would make the co-efficient (|DE(d_1,...,d_n)|+|DO(d_1,...,d_n)|) as the total number of ways to create outdegree $d_i$ on a vertex is all the ways to do so with an even orientation combined with all the ways to do so with an odd orientation.
I have a feeling that I am probably missing a double counting argument? As perhaps all odd orientations are contained within even orientations? This might account for the minus sign that I'm not understanding. 
 A: I did struggle a bit with this, but if you think long enough about it, it becomes quite intuitive.  Each even $D$ contributes positively to the monomial, and each odd $D$ negatively.  Here are the details.
As explained before the Lemma, one can find $f_G$ by taking 
$$\sum_{D}\prod_{e \in E(D)} w(e)$$ where $D$ ranges over every orientation.
We'll split $D$ into degree sequences. 
For some degree sequence $d_1, \ldots, d_n$, denote $D(d_1, \ldots, d_n)$ the set of orientations that have the corresponding outdegrees.
Then 
$$f_G = \sum_{d_1, \ldots, d_n  \geq 0} \sum_{D \in D(d_1, \ldots, d_n)} \prod_{e \in E(D)} w(e)$$
Now for some $D \in D(d_1, \ldots, d_n)$, the product $\prod_{e \in E(D)}w(e)$ 
has only two possible values, that is $\pm \prod_{i = 1}^n x_i^{d_i}$ (that's from the definition of $w(e)$).
The sign depends on the parity of $D$.  Positive if $D$ is even, negative otherwise.  Thus you could write
$$f_G = \sum_{d_1, \ldots, d_n \geq 0} \left( \sum_{D \in DE(d_1, \ldots, d_n)} \prod_{i = 1}^n x_i^{d_i} + \sum_{D \in DO(d_1, \ldots, d_n)} -\prod_{i = 1}^n x_i^{d_i}  \right)$$
The Lemma follows.
