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n-n-1-n-2-n-3-n-r-1-




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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