Given $A,B \in \mathbb{Z}\_+$ and $ 0 < t, q< 1$, I'd like to compute the coefficients $c\_n(q,A,B)$ in the expansion of the product $$\prod\_{i=0}^{A-1} \prod\_{j=0}^{B-1} \frac{1}{1-t q^{i+j}} = \sum\_{n=0}^{\infty} c\_n t^n.$$ As $q \rightarrow 1$, this returns the well known formula $$\frac{1}{(1-t)^{AB}} = \sum\_{n=0}^{\infty} \binom{AB+n-1}{n} t^n$$ which has a quick enumerative proof.


So far I've determined that the highest power of $q$ in $c\_n(q,A,B)$ is $(A-1)(B-1)n$ less than the highest power of $q$ in the $q$-binomial coefficient $\binom{AB+n-1}{n}\_q$. Can anyone see what these $c\_n$ count in the expansion of the product? Any help would be greatly appreciated!



Taylor expansion of a q-analog of the negative binomial distribution
Make the (harmless) substitution of $qt$ for $t$ and rewrite the product
as $\prod\_{i=1}^A\prod\_{j=1}^B (1-tq^{i+j-1})^{-1}$. The coefficient of 
$t^k$ is then $\sum\_n T\_{ABk}(n)q^n$, where $T\_{ABk}(n)$ is the
number of plane partitions of $n$ with at most $A$ rows, at most
$B$ columns, and trace (sum of main diagonal elements) $k$. See
*Enumerative Combinatorics*, vol. 2, Theorem 7.20.1.


