Question

If ${}^n{C}_8={}^n{C}_6,$ find ${}^n{C}_3.$

Answer 1

We know that ${}^n{C}_r={}^n{C}_q$ and $p\ne q\Rightarrow n=p+q.$

$\therefore {}^n{C}_8={}^n{C}_6\Rightarrow n=(8+6)=14.$

$\therefore {}^n{C}_3={}^{14}{C}_3=\frac{14\times 13\times 12}{3!}=\frac{14\times 13\times 12}{3\times 2\times 1}=364.$