The discussion confirms that the combinatorial identity presented, \(\sum^{r-1}_{k=0} \binom{n-1}{r-(k+1)} \times \binom{r}{k} = \binom{n+r-1}{r-1}\), is indeed Vandermonde's identity. Participants clarified that by substituting \(m = r\) and adjusting the index \(k\), the identity can be derived directly from the established formula. The identity is useful for counting distinct nonnegative integer-valued vectors that satisfy the equation \(x_{1}+x_{2}+\ldots+x_{r}=n\).
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CompuChip said:Yes, it is called http://en.wikipedia.org/wiki/Vandermonde's_identity]Vandermonde's[/PLAIN] identity.
If you plug in m = r in the first formula given in that link, and shift k ==> k - 1, you get exactly what you wrote.