The $q$-analog of the binomial coefficient $\binom{n}{k}$ may be defined as the coefficient of $x^k$ in $\prod_{i=0}^{n-1}(1+q^ix)$.
Classical arithmetic identities tend to have $q$-analogs. I am interested in the following identity: $$\binom{n}{a}\binom{n}{b}=\sum_{i=0}^{\mathrm{min}(a,b)}\binom{a+b-i}{i,a-i,b-i}\binom{n}{a+b-i}$$
This can be proved by comparing the coefficient of $x^ay^b$ in both $(1+x)^n(1+y)^n$ and $(1+x+y+xy)^n$.
Question: Is there a $q$-analog of this identity?
For example, we have $\binom{n}{1}^2=2\binom{n}{2}+n$, and its $q$-analog is $$\binom{n}{1}_q^2=(q+q^2)\binom{n}{2}_q+q^{n-1}\binom{n}{1}_q$$
What replaces $\binom{a+b-i}{i,a-i,b-i}$ in general? Is there a version where the coefficients don't depend on $n$?
Just to expand on Nate's answer (thank you!) and close the question: The $q$-analog of the identity is: $$\binom{n}{a}_q\binom{n}{b}_q=\sum_{i=0}^{\mathrm{min}(a,b)}q^{(a-i)(b-i)}\binom{a+b-i}{i,a-i,b-i}_q\binom{n}{a+b-i}_q$$