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Topic: Probability and Combinatorics
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Posts: 2
Registered: 12/20/08
Probability and Combinatorics
Posted: Dec 20, 2008 8:04 PM
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Does anybody know how to solve the following problem:


Well, actually i know that the solution is:


but i cannot prove it (mathematica calculates the first sum and gives the answer).

Maybe someone knows the simple (combinatorial?) solution?

The original problem was:

Let [tex]R ^{M} _{P} = \sum_{s = 0}^{P} {M + 1 \choose s}[/tex], for [tex]0 \leqslant P \leqslant M[/tex], [tex]P,M\in \mathbb{N}[/tex]

Proove that:
[tex]\sum_{q = 0}^{M}R^{M}_{q}\cdot R^{M}_{M - q} = (2M + 1) {2M \choose M}[/tex]

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