As usual, denote $[n]_q=1+q+\cdots+q^{n-1}$ and $[n]_q!=[1]_q[2]_q\cdots[n]_q$. Furthermore, we write
$$\binom{n}k_q=\frac{[n]_q!}{[k]_q!\cdot[n-k]_q!}.$$
As a follow up on [this MO question][1], I propose a $q$-analogue identity.

>**Question.** Can you show that
$$\sum_{k=0}^nq^{(y-n+1)k}\binom{x+k}k_q\binom{y-k}{n-k}_q
=\sum_{k=0}^nq^{n-k}\binom{x+y-k}{n-k}_q\,\,\,?$$

It would be great if we can see alternative proofs? I've a bias for combinatorial arguments. :-)


[1]: https://mathoverflow.net/questions/270115/an-interesting-identity-in-search-of-a-proof-part-i