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Question Number 41642 by Tawa1 last updated on 10/Aug/18

n(n − 1)(n − 2)(n − 3) .... (n − r + 1) = ??

$$\mathrm{n}\left(\mathrm{n}\:−\:\mathrm{1}\right)\left(\mathrm{n}\:−\:\mathrm{2}\right)\left(\mathrm{n}\:−\:\mathrm{3}\right)\:....\:\left(\mathrm{n}\:−\:\mathrm{r}\:+\:\mathrm{1}\right)\:=\:?? \\ $$

Answered by alex041103 last updated on 10/Aug/18

=((n!)/((n−r)!))

$$=\frac{{n}!}{\left({n}−{r}\right)!} \\ $$

Commented by Tawa1 last updated on 10/Aug/18

please workings

$$\mathrm{please}\:\mathrm{workings} \\ $$

Commented by alex041103 last updated on 10/Aug/18

n(n−1)...(n−r+1)=  =Π_(k=n−r+1) ^n k=Π_(k=n−r+1) ^n k ((Π_(k=1) ^(n−r) k)/(Π_(k=1) ^(n−r) k))=  =((Π_(k=1) ^n k)/(Π_(k=1) ^(n−r) k))=((n!)/((n−r)!))

$${n}\left({n}−\mathrm{1}\right)...\left({n}−{r}+\mathrm{1}\right)= \\ $$$$=\underset{{k}={n}−{r}+\mathrm{1}} {\overset{{n}} {\prod}}{k}=\underset{{k}={n}−{r}+\mathrm{1}} {\overset{{n}} {\prod}}{k}\:\frac{\underset{{k}=\mathrm{1}} {\overset{{n}−{r}} {\prod}}{k}}{\underset{{k}=\mathrm{1}} {\overset{{n}−{r}} {\prod}}{k}}= \\ $$$$=\frac{\underset{{k}=\mathrm{1}} {\overset{{n}} {\prod}}{k}}{\underset{{k}=\mathrm{1}} {\overset{{n}−{r}} {\prod}}{k}}=\frac{{n}!}{\left({n}−{r}\right)!} \\ $$

Commented by Tawa1 last updated on 10/Aug/18

God bless you sir

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$

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