Question Number 85228 by mathocean1 last updated on 20/Mar/20
Commented by mathocean1 last updated on 20/Mar/20
$${please}\:{help}\:{me}... \\ $$
Commented by abdomathmax last updated on 20/Mar/20
$${C}_{{n}} ^{{p}} −{C}_{{n}−\mathrm{1}} ^{{p}} \:=\frac{{n}!}{{p}!\left({n}−{p}\right)!}−\frac{\left({n}−\mathrm{1}\right)!}{{p}!\left({n}−{p}−\mathrm{1}\right)!} \\ $$$$=\frac{{n}\left({n}−\mathrm{1}\right)!}{{p}\left({p}−\mathrm{1}\right)!\left({n}−{p}\right)!}−\frac{\left({n}−\mathrm{1}\right)!\left({n}−{p}\right)}{{p}\left({p}−\mathrm{1}\right)!\left({n}−{p}\right)!} \\ $$$$=\frac{\left({n}−\mathrm{1}\right)!}{\left({p}−\mathrm{1}\right)!\left({n}−{p}\right)!}\left\{\frac{{n}}{{p}}−\frac{{n}−{p}}{{p}}\right\} \\ $$$$=\frac{\left({n}−\mathrm{1}\right)!}{\left({p}−\mathrm{1}\right)!\left({n}−{p}\right)!}×\mathrm{1} \\ $$$$={C}_{{n}−\mathrm{1}} ^{{p}−\mathrm{1}} \:\Rightarrow\:{C}_{{n}} ^{{p}} \:={C}_{{n}−\mathrm{1}} ^{{p}} \:+{C}_{{n}−\mathrm{1}} ^{{p}−\mathrm{1}} \\ $$