Question

$\text{if}{}^n{C}_8={}^n{C}_6,\text{then}\text{find}{}^n{C}_2$

Answer 1

Given ${}^n{C}_8{=}^n{C}_6$

$\Rightarrow \frac{n!}{8!(n-8)!}=\frac{n!}{6!(n-6)!}$

$\Rightarrow \frac{1}{8.7}=\frac{1}{(n-6)(n-7)}$

$\Rightarrow (n-6)(n-7)=56$

$\Rightarrow n^2-13n-14=0$

$\Rightarrow n=-1$ or $14$

$n$ cannot be $(-)ve,\therefore n=14$

Now ${}^n{C}_2=\frac{14!}{2!\times 12!}=\frac{14\times 13}{2}=7\times 13=91$