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Question-71550




Question Number 71550 by mhmd last updated on 17/Oct/19
Answered by MJS last updated on 17/Oct/19
it′ wrong, it must be   (((n+1)),(k) ) = ((n),(k) ) + ((n),((k−1)) )   ((n),(k) ) + ((n),((k−1)) ) =((n!)/(k!(n−k)!))+((n!)/((k−1)!(n−k+1)!))=  =((n!(n−k+1))/((k−1)!(n−k+1)!k))+((n!k)/((k−1)!(n−k+1)!k))=  =((n!(n−k+1)+n!k)/((k−1)!(n−k+1)!k))=((n!(n+1))/(k!(n+1−k)!))=  =(((n+1)!)/(k!(n+1−k)!))= (((n+1)),(k) )
$$\mathrm{it}'\:\mathrm{wrong},\:\mathrm{it}\:\mathrm{must}\:\mathrm{be} \\ $$$$\begin{pmatrix}{{n}+\mathrm{1}}\\{{k}}\end{pmatrix}\:=\begin{pmatrix}{{n}}\\{{k}}\end{pmatrix}\:+\begin{pmatrix}{{n}}\\{{k}−\mathrm{1}}\end{pmatrix} \\ $$$$\begin{pmatrix}{{n}}\\{{k}}\end{pmatrix}\:+\begin{pmatrix}{{n}}\\{{k}−\mathrm{1}}\end{pmatrix}\:=\frac{{n}!}{{k}!\left({n}−{k}\right)!}+\frac{{n}!}{\left({k}−\mathrm{1}\right)!\left({n}−{k}+\mathrm{1}\right)!}= \\ $$$$=\frac{{n}!\left({n}−{k}+\mathrm{1}\right)}{\left({k}−\mathrm{1}\right)!\left({n}−{k}+\mathrm{1}\right)!{k}}+\frac{{n}!{k}}{\left({k}−\mathrm{1}\right)!\left({n}−{k}+\mathrm{1}\right)!{k}}= \\ $$$$=\frac{{n}!\left({n}−{k}+\mathrm{1}\right)+{n}!{k}}{\left({k}−\mathrm{1}\right)!\left({n}−{k}+\mathrm{1}\right)!{k}}=\frac{{n}!\left({n}+\mathrm{1}\right)}{{k}!\left({n}+\mathrm{1}−{k}\right)!}= \\ $$$$=\frac{\left({n}+\mathrm{1}\right)!}{{k}!\left({n}+\mathrm{1}−{k}\right)!}=\begin{pmatrix}{{n}+\mathrm{1}}\\{{k}}\end{pmatrix} \\ $$

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