I am trying to enumerate a certain quantity and at some point I get the following sum:
\begin{equation} \sum_{i=0}^{m}{m \brack i}_q \sum_{j=0}^{n-m} q^{j(m-i)}{n-m \brack j}_q \sum_{k=0}^{r} {r\brack k}_q q^{(j+i)\left(r-k\right)}. \end{equation}
I wonder if this can be simplified somehow by using a $q$-binomial identity, however, I don't have any working knowledge in that area.
Any help would be much appreciated.
Partial Answer If you let $q = 1$, do the sum over $k$ and then the sum over $j$: $$ \sum_{i=0}^{m} \binom{m}{i} \sum_{j=0}^{n-m} \binom{n-m}{j} \sum_{k=0}^{r} \binom{r}{k} = 2^r \sum_{i=0}^{m} \binom{m}{i} \sum_{j=0}^{n-m} \binom{n-m}{j} = 2^r \sum_{i=0}^{m} \binom{m}{i} 2^{n-m} $$
Pull the $2^n$ outside of the summation, and the last sum is the total probability of the binomial distribution.
$$ 2^r \sum_{i=0}^{m} \binom{m}{i} 2^{n-m} = 2^{n+r} \sum_{i=0}^{m} \binom{m}{i} 2^{-m} = 2^{n+r}$$
So we learned this is a type of "iterated summation" and the limiting value is $2^{n+r}$.
In your case you need moments of the $q$-binomial distribution. Binomial theorem says:
$$ \sum_{k=0}^{r} {r\brack k}_q x^k q^{\binom{k}{2}} = \prod_{j=0}^r (1+xq^i) $$
In your case, there's no $q^{\binom{k}{2}}$ factor, the above formula does not apply.