# Taylor expansion of a q-analog of the negative binomial distribution

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!

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Not sure if this helps but in the case where A and B are equal then one can put $x_i=\sqrt{t}q^j$, $y_j=\sqrt{t}q^j$ and the expression looks like the Cauchy determinant, and can be expanded as sum of product of Schur functions. Then you are trying to compute a specialization of Schur functions labeled by a partition $\lambda$ with the alphabets being $\sqrt{t}q^i$. The quotient of alternants formula for Schur function might come handy. – Vasu vineet Nov 16 '11 at 1:47
Okay, I see the question got answered! – Vasu vineet Nov 16 '11 at 1:48

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.