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Re: Sum of squares of binomial coefficients
Posted:
Oct 12, 2012 6:13 PM
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In message <121020120727017676%edgar@math.ohio-state.edu.invalid>
Jérôme Collet <Jerome.Collet@laposte.net> wrote:
> So my question is
> \sum_{r,s}{ (\binom{r+s}{r} \binom{2m-r-s}{m-r})^2 }
> seems to be equivalent to
> \sqrt{2\pi m} \binom{2m}{m}^2
> How can I prove it ?
A bit more attention; maybe not very useful. Suppose you have
a 2m-element set. Call a quadruple of subsets (A,B,U,V) such that
1) #U = #V = m
2) #A = #B
3) #(A\cap U) = #(B\cap V)
"jolly". Then I believe that
\sum_{r,s}{ (\binom{r+s}{r} \binom{2m-r-s}{m-r})^2 }
is the number of jolly quadruples. Now \binom{2m}{m}^2 is
the number of ways of choosing pairs of subsets (U,V) such that
#U = #V = m. I am sceptical that all the rest of the jollity
counts asymptotically for \sqrt{2\pi m}. I would have expected something
a lot bigger, but I cannot give you any good reasons right now.
--
Gavin Wraith (gavin@wra1th.plus.com)
Home page: http://www.wra1th.plus.com/
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