An interesting combinatorial relation

By ajb on November 17, 2007 4:48 PM | Permalink | Comments (0) | TrackBacks (0)

$\left( \begin{array}{c} n \\ 2k+1 \end{array} ight) = \sum_{i=k+1}^{n-k} \left( \begin{array}{c} i-1 \\ k \end{array} ight) \left( \begin{array}{c} n-i \\ k \end{array} ight)$

The proof is left as an exercise.

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