Question

If ${}^{18}C_{15}+2\left({}^{18}C_{16}\right)+{}^{17}C_{16}+1={}^{n}C_3$, then $n$ is equal to

• A. $19$
• B. $20$
• C. $8$
• D. $24$
• E. $21$

Answer 1

The correct option is B

$20$

Explanation for the correct option:

Find the value of $n$:

Given, ${}^{18}C_{15}+2\left({}^{18}C_{16}\right)+{}^{17}C_{16}+1={}^{n}C_3$

$\Rightarrow$ ${}^{18}C_{15}+2\left({}^{18}C_{16}\right)+{}^{17}C_{16}+{}^{17}C_{17}={}^{n}C_3$ ; $\left[\mathbf{\because }{}^{\mathbf{n}}\mathit{C}_{\mathit{r}}\mathbf{}\mathbf{+}\mathbf{}{}^{\mathbf{n}}\mathit{C}_{\mathbf{r}\mathbf{-}\mathbf{1}}\mathbf{}\mathbf{=}\mathbf{}{}^{\mathbf{n}\mathbf{+}\mathbf{1}}\mathit{C}_{\mathbf{r}}\right]$

$\Rightarrow$ ${}^{18}C_{15}+2\left({}^{18}C_{16}\right)+{}^{18}C_{17}={}^{n}C_3$

$\Rightarrow$$\left({}^{18}C_{15}+{}^{18}C_{16}\right)+\left({}^{18}C_{16}+{}^{18}C_{17}\right)={}^{n}C_3$

$\Rightarrow$ ${}^{19}C_{16}+{}^{19}C_{17}={}^{n}C_3$

$\Rightarrow$ ${}^{20}C_{17}={}^{n}C_3$

$\mathbf{\therefore }\mathit{n}\mathbf{}\mathbf{=}\mathbf{}\mathbf{20}$

Hence, Option ‘B’ is Correct.