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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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    $\begingroup$ 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? $\endgroup$
    – Somnath Basu
    Commented Oct 23, 2011 at 5:12
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    $\begingroup$ I am pretty sure this is a hw question as I could do it in 5 minutes. $\endgroup$
    – John Jiang
    Commented Oct 23, 2011 at 6:02

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The generating function in $n$ is $((1+t)^2+1)^m$. The case m=n is http://oeis.org/A006139.

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  • $\begingroup$ ..I found it had no beautiful solution...but thx.. $\endgroup$
    – user18717
    Commented Oct 28, 2011 at 15:46

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