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

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

$$\left(\begin{array}{c}\mathrm{n}+1\\ \mathrm{k}+1\end{array}\right)=\frac{(\mathrm{n}+1)!}{(\mathrm{k}+1)!(\mathrm{n}-\mathrm{k})!}$$$$\left(\begin{array}{c}\mathrm{n}\\ \mathrm{k}\end{array}\right)=\frac{\mathrm{n}!}{\mathrm{k}!(\mathrm{n}-\mathrm{k})!}$$$$\left(\begin{array}{c}\mathrm{n}\\ \mathrm{k}+1\end{array}\right)=\frac{\mathrm{n}!}{(\mathrm{k}+1)!(\mathrm{n}-\mathrm{k}-1)!}=\frac{\mathrm{n}!(\mathrm{n}-\mathrm{k})}{(\mathrm{k}+1)!(\mathrm{n}-\mathrm{k})!}$$$$\left(\begin{array}{c}\mathrm{n}\\ \mathrm{k}\end{array}\right)+\left(\begin{array}{c}\mathrm{n}\\ \mathrm{k}+1\end{array}\right)=\frac{\mathrm{n}!}{(\mathrm{n}-\mathrm{k})!}(\frac{1}{\mathrm{k}!}+\frac{\mathrm{n}-\mathrm{k}}{(\mathrm{k}+1)!})=\frac{\mathrm{n}!}{(\mathrm{n}-\mathrm{k})!(\mathrm{k}+1)!}(\mathrm{k}+1+\mathrm{n}-\mathrm{k})$$$$=\frac{(\mathrm{n}+1)!}{(\mathrm{n}-\mathrm{k})!(\mathrm{k}+1)!}=\left(\begin{array}{c}\mathrm{n}+1\\ \mathrm{k}+1\end{array}\right)$$

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