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Question Number 74337 by Maclaurin Stickker last updated on 22/Nov/19

$$findthecontractedformof:$$$$\left(\begin{array}{c}n\\ p\end{array}\right)+2\left(\begin{array}{c}n\\ p+1\end{array}\right)+\left(\begin{array}{c}n\\ p+2\end{array}\right)$$

Answered by MJS last updated on 23/Nov/19

$$\left(\begin{array}{c}n\\ p\end{array}\right)=\frac{n!}{p!(n-p)!}=\frac{n!(p+1)(p+2)}{p!(n-p)!(p+1)(p+2)}$$$$2\left(\begin{array}{c}n\\ p+1\end{array}\right)=\frac{2n!}{(p+1)!(n-p-1)!}=\frac{2n!(n-p)}{p!(n-p)!(p+1)}=$$$$=\frac{2n!(n-p)(p+2)}{p!(n-p)!(p+1)(p+2)}$$$$\left(\begin{array}{c}n\\ p+2\end{array}\right)=\frac{n!}{(p+2)!(n-p-2)!}=$$$$=\frac{n!(n-p)(n-p-1)}{p!(n-p)!(p+1)(p+2)}$$$$$$$$\left(\begin{array}{c}n\\ p\end{array}\right)+2\left(\begin{array}{c}n\\ p+1\end{array}\right)+\left(\begin{array}{c}n\\ p+2\end{array}\right)=$$$$=\frac{n!((p+1)(p+2)+2(n-p)(p+2)+(n-p)(n-p-1))}{p!(n-p)!(p+1)(p+2)}=$$$$=\frac{n!(n^2+3n+2)}{p!(n-p)!(p+1)(p+2)}=$$$$=\frac{n!(n+1)(n+2)}{p!(n-p)!(p+1)(p+2)}=\frac{(n+2)!}{(p+2)!(n-p)!}=$$$$=\frac{(n+2)!}{(p+2)!((n+2)-(p+2))!}=\left(\begin{array}{c}n+2\\ p+2\end{array}\right)$$

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