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

The Correct Option is 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}. $