Question

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

Answer 1

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

if ${}^n{C}_r={}^n{C}_p$

Then either $r=p$ or $r=n-p$

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

$8=n-2$

$10=n$

$\therefore n=10$

Calculating ${}^n{C}_2$

Put $n=10$

${}^n{C}_2={}^{10}{C}_2$

$=\frac{10!}{2!(10-2)!}$

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

$=\frac{10\times 9}{2}$

$=45$

${}^n{C}_2=45$.