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$$\frac{\binom{n}{j}_q\binom{n+1}{j}_q \cdots\binom{n+k-1}{j}_q}{\binom{j}{j}_q\binom{j+1}{j}_q\cdots\binom{j+k-1}{j}_q}$$ where $n,j,k$ are non-negative integers.

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What do the $q$ subscripts on the binomial coefficients mean? –  Henning Makholm Dec 28 '11 at 22:39
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It means it is q binomial coefficient and here is the link to its definition mathworld.wolfram.com/q-BinomialCoefficient.html –  sophie668 Dec 28 '11 at 22:44
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EDIT: I previously gave a flawed counterexample to the question, but I discovered a flaw in it and now believe the statement in the question to be true. The last two paragraphs are not a formal proof of this, but I believe could be turned into such a proof.

If $n<j$ then this is $0$, which is certainly a polynomial in $q$. Otherwise, we have $$\frac{\binom{n}{j}_q\binom{n+1}{j}_q \cdots\binom{n+k-1}{j}_q}{\binom{j}{j}_q\binom{j+1}{j}_q\cdots\binom{j+k-1}{j}_q} $$ $$= \frac{([n]_q[n-1]_q\cdots[n-j+1]_q)\cdots([n+k-1]_q[n+k-2]_q\cdots[n+k-j]_q)}{([j]_q[j-1]_q\cdots[1]_q)\cdots([j+k-1]_q[j+k-2]_q\cdots[k]_q)}$$ where $[m_q] = 1 + q + \cdots + q^{m-1}=(q-\zeta_m)(q-\zeta_m^2)\cdots(q-\zeta_m^{m-1})$ and $\zeta_m$ is a primitive $m^{th}$ root of unity. These polynomials are irreducible over $\mathbb{C}$ and since $\mathbb{C}[x]$ is a unique factorization domain we can only write $$\frac{([n]_q[n-1]_q\cdots[n-j+1]_q)\cdots([n+k-1]_q[n+k-2]_q\cdots[n+k-j]_q)}{([j]_q[j-1]_q\cdots[1]_q)\cdots([j+k-1]_q[j+k-2]_q\cdots[k]_q)}$$ as a polynomial in $q$ over $\mathbb{C}$ if the factors $(q-\zeta_m^l)$ in the denominator cancel with factors in the numerator. Note that $\zeta_{im}^i = \zeta_i$ (this ignores our choice of primitive root, but we can make our choice arbitrarily).

The factor $(q-\zeta_j)$ of $[j]_q$ cancels with the factor $(q-\zeta_{im}^i)$ of $[ij]_q$ chosen as the unique multiple of $j$ among $n,n-1,\ldots,n-j+1$. Similarly, the factor $(q-\zeta_j^2)$ cancels with $(q-\zeta_{im}^{2i})$, and repeating this procedure we cancel all the factors of $[j]_q$ with factors of $[ij]_q$. We can pick a multiple $i'(j-1)$ of $j-1$ among $\{n,n-1,\ldots,n-j+1\}-\{ij\}$, and similarly a multiple of $j-2$ among $\{n,n-1,\ldots,n-j+1\}-\{ij,i'(j-1)\}$, and by repeating this procedure cancel all factors of $[j]_q[j-1]_q\cdots[1]_q$ with factors of $[n]_q[n-1]_q\cdots[n-j+1]_q$.

These are easy factors to cancel, because some multiple of $j$ is among $n+1,n,\ldots,n-j+1$. But what about the factors of $[j+1]_q$? Well, we have $i'(j+1)=i'j+i'$ among $n,n-1,\ldots,n-j+2$ for some $i'$, as the gap between $i'(j+1)$ and $(i'+1)(j+1)$ is $j+1$, and so we can cancel its factors with those of $[n-j+1]_q$ or of one of $[n+1]_q,\ldots,[n-j+2]_q$. But wait, we've already used $[n-j+1]_q$ to cancel factors in the denominator! True, but if $n-j+1$ is a multiple of $j+1$ then we must have used it to cancel factors of $[\frac{j+1}{2}]_q$ or smaller, so we can still cancel at least half the factors of $[j+1]_q$, and the other half will involve roots of unity of smaller order and so can be canceled with other leftovers among $[n]_q,\ldots,[n-j+1]_q$. The same should extend to $[j]_q,[j-1]_q,\ldots,[2]_q$. Iterating this argument should allow cancellation of all the factors in the denominator.

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