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

$$\mathrm{n}(\mathrm{n}-1)(\mathrm{n}-2)(\mathrm{n}-3)....(\mathrm{n}-\mathrm{r}+1)=??$$

Answered by alex041103 last updated on 10/Aug/18

$$=\frac{n!}{(n-r)!}$$

Commented by Tawa1 last updated on 10/Aug/18

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

Commented by alex041103 last updated on 10/Aug/18

$$n(n-1)...(n-r+1)=$$$$=\underset{k=n-r+1}{\prod ^n}k=\underset{k=n-r+1}{\prod ^n}k\frac{\underset{k=1}{\prod ^{n-r}}k}{\underset{k=1}{\prod ^{n-r}}k}=$$$$=\frac{\underset{k=1}{\prod ^n}k}{\underset{k=1}{\prod ^{n-r}}k}=\frac{n!}{(n-r)!}$$

Commented by Tawa1 last updated on 10/Aug/18

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

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