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I'm trying to give a combinatorial proof of the q-Chu-Vandermonde identity: $\binom{m + n}{k}_{q} =\sum_{j} \binom{m}{k - j}_{q} \binom{n}{j}_{q} q^{j(m-k+j)}.$

I understand that the LHS counts the number of dimension $k$ subspaces of a vector space of dimension $m + n$ of a Finite Field with $q$ elements. I see how the traditional committee forming argument for the Chu-Vandermonde identity could be applied since the RHS is the sum over all possible combinations of dimension $j$ subspaces of a vector space of size $n$ and dimension $k-j$ subspaces of a vector space of size $m$, but I am not sure how to make this proof more rigorous specifically regarding the algebra aspect.

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  • $\begingroup$ Fix an exact sequence $0 \to M \to P \to N \to 0$ of vector spaces, where $M$, $P$ and $N$ have dimensions $m$, $m+n$ and $n$, respectively. If $U$ is any $k$-dimensional subspace of $P$, then the dimension of the projection of $U$ onto $N$ and the dimension of the preimage of $U$ under $M \to P$ add up to $k$. For any pair $\left(A, B\right)$ of a $j$-dimensional subspace $A$ of $N$ and a $k-j$-dimensional subspace $B$ of $M$, there are exactly $q^{j\left(m-k+j\right)}$ subspaces of $P$ (all $k$-dimensional) whose projection onto $N$ is $A$ and whose preimage of $U$ under $M \to P$ is $B$. $\endgroup$ – darij grinberg Oct 5 '17 at 2:09
  • $\begingroup$ NB: Take the above with a grain of salt; not actually proven. $\endgroup$ – darij grinberg Oct 5 '17 at 2:22
  • $\begingroup$ @darijgrinberg I'm not exactly sure how to see the fact that there are $q^{(j)(m-k+j)}$ subspaces of P which satisfy the criteria. $\endgroup$ – Harry Xi Oct 5 '17 at 3:31

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