Solveeit Logo
Login

Question

Mathematics Question on Combinations

nCr+2(nCr−1)+nCr−2^{n}C_{r} + 2( {^{n}C_{r-1} }) + {^{n}C_{r-2}} is equal to:

A

n+2Cr{^{n +2}C_r}

B

nCr+1{^{n }C_{r+1}}

C

n−1Cr+1{^{n - 1 }C_{r+1}}

D

none of these

Answer

n+2Cr{^{n +2}C_r}

Explanation

Solution

Let A=nCr+2(nCr−1)+nCr−2A= ^{n}C_{r} + 2(^{n}C_{r-1}) + ^{n}C_{r-2} =n!r!(n−r)!+2n!(r−1)!(n−r+1)!+n!(r−2)!(n−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)!} =n![(n−r+2.n−r+1)+2(n−r+2)r+r(r−1)]r!(n−r+2)!= \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)!} =[n![(n2−nr+n−nr+r2−r+2n−2r+2+2nr−2r2+4r+r2−r)]]r!(n−r+2)! = \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)!} =(n2+3n+2)n!r!(n−r+2)!=(n+1)(n+2)n!r!(n−r+2)!= \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)!} =(n+2)!r!(n+2−r)!=n+2Cr.= \frac{\left(n+2\right)!}{r!\left(n+2-r\right)!} = ^{n+2}C_{r}.