Question-115683

Question Number 115683 by mohammad17 last updated on 27/Sep/20

Answered by Dwaipayan Shikari last updated on 27/Sep/20

(((n+1)),((k+1)) )=(((n+1)!)/((k+1)!(n−k)!)) ((n),(k) )=((n!)/(k!(n−k)!)) ((n),((k+1)) )=((n!)/((k+1)!(n−k−1)!))=((n!(n−k))/((k+1)!(n−k)!)) ((n),(k) )+ ((n),((k+1)) )=((n!)/((n−k)!))((1/(k!))+((n−k)/((k+1)!)))=((n!)/((n−k)!(k+1)!))(k+1+n−k) =(((n+1)!)/((n−k)!(k+1)!))= (((n+1)),((k+1)) )

$$\begin{pmatrix}{\mathrm{n}+\mathrm{1}}\\{\mathrm{k}+\mathrm{1}}\end{pmatrix}=\frac{\left(\mathrm{n}+\mathrm{1}\right)!}{\left(\mathrm{k}+\mathrm{1}\right)!\left(\mathrm{n}−\mathrm{k}\right)!} \\ $$$$\begin{pmatrix}{\mathrm{n}}\\{\mathrm{k}}\end{pmatrix}=\frac{\mathrm{n}!}{\mathrm{k}!\left(\mathrm{n}−\mathrm{k}\right)!} \\ $$$$\begin{pmatrix}{\mathrm{n}}\\{\mathrm{k}+\mathrm{1}}\end{pmatrix}=\frac{\mathrm{n}!}{\left(\mathrm{k}+\mathrm{1}\right)!\left(\mathrm{n}−\mathrm{k}−\mathrm{1}\right)!}=\frac{\mathrm{n}!\left(\mathrm{n}−\mathrm{k}\right)}{\left(\mathrm{k}+\mathrm{1}\right)!\left(\mathrm{n}−\mathrm{k}\right)!} \\ $$$$\begin{pmatrix}{\mathrm{n}}\\{\mathrm{k}}\end{pmatrix}+\begin{pmatrix}{\mathrm{n}}\\{\mathrm{k}+\mathrm{1}}\end{pmatrix}=\frac{\mathrm{n}!}{\left(\mathrm{n}−\mathrm{k}\right)!}\left(\frac{\mathrm{1}}{\mathrm{k}!}+\frac{\mathrm{n}−\mathrm{k}}{\left(\mathrm{k}+\mathrm{1}\right)!}\right)=\frac{\mathrm{n}!}{\left(\mathrm{n}−\mathrm{k}\right)!\left(\mathrm{k}+\mathrm{1}\right)!}\left(\mathrm{k}+\mathrm{1}+\mathrm{n}−\mathrm{k}\right) \\ $$$$=\frac{\left(\mathrm{n}+\mathrm{1}\right)!}{\left(\mathrm{n}−\mathrm{k}\right)!\left(\mathrm{k}+\mathrm{1}\right)!}=\begin{pmatrix}{\mathrm{n}+\mathrm{1}}\\{\mathrm{k}+\mathrm{1}}\end{pmatrix} \\ $$

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