Question Number 769 by rishabh last updated on 09/Mar/15
$${Prove}\:{that}\:{product}\:{of}\:{any}\:\:{n}\:{consecutive}\:{integers} \\ $$$${is}\:{divisiblr}\:{by}\:{n}! \\ $$
Commented by 123456 last updated on 09/Mar/15
$${n}\left({n}+\mathrm{1}\right)\centerdot\centerdot\centerdot\left(\mathrm{2}{n}−\mathrm{1}\right) \\ $$
Answered by prakash jain last updated on 09/Mar/15
![n consecutive number ending with p are (P) P=p(p−1)...(p−n+1) (P/(n!)) = ((p(p−1)...(p−n+1))/(n!)) =(([p(p−1)...(p−n+1)][(p−n)(p−n−1)...1])/(n!)) =((p!)/(n!(p−n)!))=^p C_n ^p C_n is always an integer so P is divisible by n!.](https://www.tinkutara.com/question/Q771.png)
$${n}\:\mathrm{consecutive}\:\mathrm{number}\:\mathrm{ending}\:\mathrm{with}\:{p}\:\mathrm{are}\:\left({P}\right) \\ $$$${P}={p}\left({p}−\mathrm{1}\right)…\left({p}−{n}+\mathrm{1}\right) \\ $$$$\frac{{P}}{{n}!}\:=\:\frac{{p}\left({p}−\mathrm{1}\right)…\left({p}−{n}+\mathrm{1}\right)}{{n}!} \\ $$$$=\frac{\left[{p}\left({p}−\mathrm{1}\right)…\left({p}−{n}+\mathrm{1}\right)\right]\left[\left({p}−{n}\right)\left({p}−{n}−\mathrm{1}\right)…\mathrm{1}\right]}{{n}!} \\ $$$$=\frac{{p}!}{{n}!\left({p}−{n}\right)!}=\:^{{p}} {C}_{\mathrm{n}} \: \\ $$$$\:^{{p}} {C}_{{n}} \:\mathrm{is}\:\mathrm{always}\:\mathrm{an}\:\mathrm{integer}\:\mathrm{so}\:{P}\:\mathrm{is}\:\mathrm{divisible}\:\mathrm{by}\:{n}!. \\ $$