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quasar987

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*Show that there are [itex]\binom{r}{k}\binom{n-1}{n-r-k}[/itex] solutions to the equation [itex]x_1+...+x_r=n[/itex] for which exactly k of the r terms of the sum are nul.*

There are [itex]\binom{r}{k}[/itex] ways of choosing which k of the r [itex]x_i[/itex]'s are zero, and there are

[tex]\binom{n+(r-k)-1}{n}[/tex]

distinc solutions to the resulting equation. What is wrong with that? If nothing, how are the two binomial coefficients equal?