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Bijective proof of a combinatorial identity: $\sum\limits_{k=0}^n\binom nk^2 \binom k{n-m}=\binom nm \binom{n+m}m$
Identity
\begin{equation}
\sum_{k=0}^n\binom nk^2 \binom k{n-m}=\binom nm \binom{n+m}m \tag{1}
\end{equation}
was used in an answer here. As shown in that answer, (1) easily reduces to
\begin{...
- 128k