Question

If ${}^{n}C_r$ denotes the number of combinations of $n$ things taken $r$ at a time, then the expression ${}^{n}C_{r+1}+{}^{n}C_{r-1}+2.{}^{n}C_r$ equals?

• A. ${}^{n+2}C_r$
• B. ${}^{n+2}C_{r+1}$
• C. ${}^{n+1}C_r$
• D. ${}^{n+1}C_{r+1}$

Answer 1

The correct option is B

${}^{n+2}C_{r+1}$

Explanation for the correct option:

Given that ${}^{n}C_r$ denotes the number of combinations of $n$ things taken $r$ at a time,

We know that ${}^{n}C_{r-1}+{}^{n}C_r={}^{n+1}C_r$

$\begin{array}{ccclc}\therefore & {}^{n}C_{r+1}+{}^{n}C_{r-1}+2.{}^{n}C_r& =& {}^{n}C_{r+1}+{}^{n}C_{r-1}+{}^{n}C_r+{}^{n}C_r& \\ & & =& \left({}^{n}C_{r+1}+{}^{n}C_r\right)+\left({}^{n}C_{r-1}+{}^{n}C_r\right)& \\ & & =& \left({}^{n+1}C_{r+1}\right)+\left({}^{n+1}C_r\right)& \left[\because {}^n{C}_m{+}^n{C}_{m-1}={}^{n+1}{C}_m\right]\\ \Rightarrow & {}^{n}C_{r+1}+{}^{n}C_{r-1}+2.{}^{n}C_r& =& {}^{n+2}C_{r+1}& \left[\because {}^n{C}_m{+}^n{C}_{m-1}={}^{n+1}{C}_m\right]\end{array}$

Hence, the correct option is option(B) i.e. ${}^{n+2}C_{r+1}$