Combinatorial identity. Using echelon matrices. Determine the exponents $e_i$ s.t. the following identity is correct.
$$\sum\limits_{i=0}^k q^{e_i} {\binom mi}_q {\binom{n}{k-i}}_q = {\binom{n+m}{k}}_q$$
Note: When $q=1$ the equation reduces to a binomial coefficients identity.
Note: Consider echelon matrices.
 A: Let's count such matrices in a different way to obtain the left-hand side. Consider such a matrix $M$. 
Let's "cut" the matrix after $m$ columns, meaning $$M=\begin{pmatrix}P&\vdots&Q\end{pmatrix}$$ where $P$ is an $(m+n)\times n$ matrix and $Q$ is an $n\times (m+n)$ matrix. Then $P$ also has echelon form, of rank $i$ where $0\leq i\leq k$. Further $i\leq m$ because $P$ only has $m$ columns. This means $P$ only has $i\leq m$ non-zero rows so the last $n$ rows are all zeroes. Now let's "cut" $P$ after row $m$ and $Q$ after rows $i, n+i$:
$$P=\begin{pmatrix}A\\0\end{pmatrix}, Q=\begin{pmatrix}B\\C\\D\end{pmatrix}$$ 
$M$ is described by the $m\times m$ echelon matrix $A$ (of rank $i$), the $n\times n$ echelon matrix $C$ (of rank $k-i$) and $i\times n$ matrix $B$. The condition on the matrix $B$ is that $k-i$ of its columns (the ones above the pivot columns of $C$) are zero, 
Clearly, we can revert this procedure and assemble such matrices $A,B,C$ into an $(m+n)\times (m+n)$ echelon matrix $M$ of rank $k$. 
Now $A$ can be chosen in $\binom mi_q$ ways, $B$ in $q^{(n-k+i)i}$ ways, and $C$ in $\binom n{k-i}_q$ ways. 
This yields the required identity $$\sum_{i=0}^k q^{(n-k+i)i}\binom mi_q\binom n{k-i}_q=\binom {m+n}k_q$$ so $e_i=i(n+k-i)$. 
