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



A "quantum" identity: in search of a proof -Part II
Both sides are equal to $\binom{x+y+1}{n}\_q$. This enumerates lattice paths in an $n\times (x+y-n+1)$ rectangle, according to the area statistic. We will assume that these paths start at $(0,0)$ and end at $(x+y-n+1,n)$, and they are only directed East or North. Here are two ways to enumerate it:


**First count**: For each path there will be a unique $k$, so it has an East step of the form $(x,k)\to (x+1,k)$. The area generating function of these paths is precisely
$$q^{y-n+1}\binom{x+k}{k}\_q\binom{y-k}{n-k}\_q.$$
Now, summing over all possible $k$, we enumerate all paths in the rectangle.


**Second count**: For each path there will be a unique $k$ so that it has an East step $(x+y-n,k)\to (x+y-n+1,k)$. Similarly the generating function of the areas of these paths is
$$q^{n-k}\binom{x+y-k}{n-k}\_q$$
and just like above, summing over all possible $k$ enumerates all paths. 


