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, 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. :-)

As usual, denote $[n]_q=1+q+\cdots+q^{n-1}=\frac{\,\,1-q^n}{1-q}$ 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, 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. :-)