how do I prove this formula combinatorially?
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Proof of a combinatorial identity: $\sum\limits_{i=0}^n {2i \choose i}{2(n-i)\choose n-i} = 4^n$
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Proof through combinatorial argument
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Proving formula for product of first n odd numbers
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How to derive this Combinatorial identity
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proving combinatorics identity - $\sum_{k=0}^m{n-k \choose m-k}={n+1 \choose m}$
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Finding a combinatorial argument for an interesting identity: $\sum_k \binom nk \binom{m+k}n = \sum_i \binom ni \binom mi 2^i$