Seeking a combinatorial proof $\sum _{k=0}^{2n} (-1)^k \binom{4n-2k}{2n-k}\binom{2k}{k}=\binom{2n}{n}\times 2^{2n}$
Combinatorial proof that $\sum \limits_{k=0}^n \binom{2k}{k} \binom{2n-2k}{n-k} (-1)^k = 2^n \binom{n}{n/2}$ when $n$ is even