Question

If ${}^n{C}_{12}{=}^n{C}_8,$ then value of ${}^n{C}_{19}$ is

• A. 1
• B. 20
• C. 210
• D. 1540

Answer 1

Answer: (B)

Given, ${}^n{C}_{12}{=}^n{C}_8$
$\Rightarrow$ $\frac{n!}{(n-12)!12!}=\frac{nQ}{(n-8)!8!}$
$\Rightarrow$ $(n-8)(n-9)(n-10)(n-11)$
$=12\times 11\times 10\times 9$
$\Rightarrow$ $n-8=12\Rightarrow n=20$
$\therefore$ ${}^{20}{C}_{19}=20$