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Question Number 34878 by NECx last updated on 12/May/18

$$showthat$$$${\stackrel{n+r-1}{C}}_{n-1}={\stackrel{n+r-1}{C}}_r$$

Answered by Rasheed.Sindhi last updated on 12/May/18

$${}^{\mathrm{n}}{\mathrm{C}}_{\mathrm{r}}=\frac{\mathrm{n}!}{\mathrm{r}!(\mathrm{n}-\mathrm{r})!}$$$${\stackrel{n+r-1}{C}}_{n-1}={\stackrel{n+r-1}{C}}_r$$$$\mathrm{LHS}:\frac{(\mathrm{n}+\mathrm{r}-1)!}{(\mathrm{n}-1)!(\mathrm{n}+\mathrm{r}-1-\overline{\mathrm{n}-1})!}$$$$\frac{(\mathrm{n}+\mathrm{r}-1)!}{(\mathrm{n}-1)!(\mathrm{n}+\mathrm{r}-1-\mathrm{n}+1)!}$$$$\frac{(\mathrm{n}+\mathrm{r}-1)!}{\mathrm{r}!(\mathrm{n}-1)!}$$$$\mathrm{RHS}:\frac{(\mathrm{n}+\mathrm{r}-1)!}{\mathrm{r}!(\mathrm{n}+\mathrm{r}-1-\mathrm{r})!}$$$$\frac{(\mathrm{n}+\mathrm{r}-1)!}{\mathrm{r}!(\mathrm{n}-1)!}$$$$\mathrm{LHS}=\mathrm{RHS}$$$$\mathrm{Proved}$$

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