Assume $n$ and $m$ are positive integers.
My eventual wish is in finding a $q$-analogue of the below identity but for now I wish to see alternative proofs. I can supply a justification using the Wilf-Zeilberger methodology.
QUESTION. Can you provide a proof for the equality? $$\sum_{k=0}^{n+m} \binom{2n+2m}{2m} \binom{2n}{n+m-k} \binom{4m}{2k} =\sum_{j=0}^n \binom{4m}{2m} \binom{2n+2m}{n-j} \binom{4m+2j-1}{2j}.$$
