Question

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

• A. $19$
• B. $20$
• C. $18$
• D. $24$

Answer 1

The correct option is B

$20$

Finding the value of n:

Given,

${}^{18}{c}_{15}+{}^{18}{c}_{16}+{}^{18}{c}_{16}+{}^{17}{c}_{16}+1={}^n{c}_3$

As we know,

${}^{\mathbf{n}}{\mathbf{c}}_{\mathbf{r}\mathbf{+}\mathbf{1}}\mathbf{}\mathbf{+}\mathbf{}{}^{\mathbf{n}}{\mathbf{c}}_{\mathbf{r}}\mathbf{}\mathbf{=}\mathbf{}{}^{\mathbf{n}\mathbf{+}\mathbf{1}}{\mathbf{c}}_{\mathbf{r}\mathbf{+}\mathbf{1}}$

$\Rightarrow$${}^{19}{c}_{16}+{}^{18}{c}_{16}+{}^{17}{c}_{16}+{}^{17}{c}_{17}={}^n{c}_3$

$\Rightarrow$ ${}^{19}{c}_{16}+{}^{18}{c}_{16}+{}^{18}{c}_{17}={}^n{c}_3$

$\Rightarrow$ ${}^{19}{c}_{16}+{}^{19}{c}_{17}={}^n{c}_3$

$\Rightarrow$ ${}^{20}{c}_{17}={}^n{c}_3$

$\Rightarrow$ ${}^{20}{c}_3={}^n{c}_3$

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

Hence, option(B) is correct.