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..I wonder if the following formula can be calculated?

$ \sum_{k=0}^m {m \choose k} {2k \choose n} $

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Look at the FAQ - even if such a thing can be calculated why do we care and what is your motivation? In particular, have you looked at various methods that are already available (combinatorial methods, generating functions etc.) and see if they work? – Somnath Basu Oct 23 '11 at 5:12
I am pretty sure this is a hw question as I could do it in 5 minutes. – John Jiang Oct 23 '11 at 6:02

The generating function in $n$ is $((1+t)^2+1)^m$. The case m=n is

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..I found it had no beautiful solution...but thx.. – user18717 Oct 28 '11 at 15:46

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