Combinatorial proof of $\sum\limits_{k=0}^n \binom{2k}{k} \binom{2n-2k}{n-k} =4^n$

Asked Oct 08 '13 at 13:50

Active Oct 08 '13 at 13:57

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$$\sum_{k=0}^n \binom{2k}{k} \binom{2n-2k}{n-k} = 4^n$$

Is there a combinatorial proof of above identity, without any arithmetic transformation?
Thanks...

edited Oct 08 '13 at 13:57

Lord_Farin

asked Oct 08 '13 at 13:50

Uyas

2

http://math.stackexchange.com/questions/72367/proof-of-a-combinatorial-identity-sum-i-0n-2i-choose-i2n-i-choose-n?lq=1 – Jaakko Seppälä Oct 08 '13 at 14:13

0 Answers0