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Question Number 220876 by Spillover last updated on 20/May/25

Answered by Rasheed.Sindhi last updated on 20/May/25

(a) (n+2)(n+1)n =(((n+2)(n+1)n(n−1)!)/((n−1)!)) =(((n+2)!)/((n−1)!))

$$\left({a}\right)\:\left({n}+\mathrm{2}\right)\left({n}+\mathrm{1}\right){n} \\ $$$$\:\:\:\:\:\:=\frac{\left({n}+\mathrm{2}\right)\left({n}+\mathrm{1}\right){n}\left({n}−\mathrm{1}\right)!}{\left({n}−\mathrm{1}\right)!} \\ $$$$\:\:\:\:\:=\frac{\left({n}+\mathrm{2}\right)!}{\left({n}−\mathrm{1}\right)!} \\ $$

Commented by Spillover last updated on 20/May/25

correct

$${correct} \\ $$

Answered by Rasheed.Sindhi last updated on 21/May/25

(b) n(n−1)...(n−r+1) =((n(n−1)...(n−r+1)(n−r)!)/((n−r)!)) =((n!)/((n−r)!))

$$\left({b}\right)\:{n}\left({n}−\mathrm{1}\right)...\left({n}−{r}+\mathrm{1}\right) \\ $$$$=\frac{{n}\left({n}−\mathrm{1}\right)...\left({n}−{r}+\mathrm{1}\right)\left({n}−{r}\right)!}{\left({n}−{r}\right)!} \\ $$$$=\frac{{n}!}{\left({n}−{r}\right)!} \\ $$$$ \\ $$

Answered by Rasheed.Sindhi last updated on 21/May/25

(c) (((n+1)n(n−1))/(3×2×1)) = (((n+1)n(n−1)(n−2)!)/(3!(n−2)!)) = (((n+1)!)/(3!(n−2)!))

$$\left({c}\right)\:\frac{\left({n}+\mathrm{1}\right){n}\left({n}−\mathrm{1}\right)}{\mathrm{3}×\mathrm{2}×\mathrm{1}} \\ $$$$\:\:\:\:\:\:\:=\:\frac{\left({n}+\mathrm{1}\right){n}\left({n}−\mathrm{1}\right)\left({n}−\mathrm{2}\right)!}{\mathrm{3}!\left({n}−\mathrm{2}\right)!} \\ $$$$\:\:\:\:\:\:\:=\:\frac{\left({n}+\mathrm{1}\right)!}{\mathrm{3}!\left({n}−\mathrm{2}\right)!} \\ $$

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