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Question-85228




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

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