Question

If ${}^n{C}_8{=}^n{C}_2$, find ${}^n{C}_2$.

Answer 1

Given, ${}^n{C}_8{=}^n{C}_2$

${}^n{C}_2=\frac{n!}{2!(n-2)!}$

${}^n{C}_{n-2}=\frac{(n-2)!}{2!(n-(n-2\left)\right)!}=\frac{n!}{2!(n-2)!}$

$⟹{}^n{C}_8{=}^n{C}_2{=}^n{C}_{n-2}$

$\Rightarrow n-2=8$,$\Rightarrow n=10$

$\therefore {}^n{C}_2{=}^{10}{C}_2=\frac{10!}{2!(10-2)!}$

$=\frac{10!}{2!8!}=\frac{10\times 9\times 8!}{2\times 1\times 8!}=45$