Question

${}^n{C}_r+2({}^n{C}_{r-1})+{}^n{C}_{r-2}$ is equal to:

• A. ${}^{n+2}{C}_r$
• B. ${}^n{C}_{r+1}$
• C. ${}^{n-1}{C}_{r+1}$
• D. none of these

Answer 1

Answer: (A)

Let $A{=}^n{C}_r+2{(}^n{C}_{r-1}){+}^n{C}_{r-2}$
$=\frac{n!}{r!\left(n-r\right)!}+\frac{2n!}{\left(r-1\right)!\left(n-r+1\right)!}+\frac{n!}{\left(r-2\right)!\left(n-r+2\right)!}$
$=\frac{n!\left[\left(n-r+2.n-r+1\right)+2\left(n-r+2\right)r+r\left(r-1\right)\right]}{r!\left(n-r+2\right)!}$
$=\frac{\left[n!\left[\left(n^2-nr+n-nr+r^2-r+2n-2r+2+2nr-2r^2+4r+r^2-r\right)\right]\right]}{r!\left(n-r+2\right)!}$
$=\frac{\left(n^2+3n+2\right)n!}{r!\left(n-r+2\right)!}=\frac{\left(n+1\right)\left(n+2\right)n!}{r!\left(n-r+2\right)!}$
$=\frac{\left(n+2\right)!}{r!\left(n+2-r\right)!}{=}^{n+2}{C}_r.$