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Question Number 150554 by pete last updated on 13/Aug/21
$$\mathrm{Show}\:\mathrm{that}\:\mathrm{n}_{\mathrm{C}_{\mathrm{r}} } =\frac{\mathrm{n}\left(\mathrm{n}−\mathrm{1}\right)\left(\mathrm{n}−\mathrm{2}\right)...\left(\mathrm{n}−\mathrm{r}+\mathrm{1}\right)}{\mathrm{r}!} \\ $$
Answered by Ar Brandon last updated on 13/Aug/21
$$\frac{{n}\left({n}−\mathrm{1}\right)\left({n}−\mathrm{2}\right)...\left({n}−{r}+\mathrm{1}\right)}{{r}!} \\ $$$$=\frac{{n}\left({n}−\mathrm{1}\right)...\left({n}−{r}+\mathrm{1}\right)\left({n}−{r}\right)!}{\left({n}−{r}\right)!{r}!} \\ $$$$=\frac{{n}!}{\left({n}−{r}\right)!{r}!}=\overset{{n}} {\:}\mathrm{C}_{{r}} \\ $$
Commented by pete last updated on 15/Aug/21
$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{sir} \\ $$