Question

Determine n if

$\left(i\right){}^{2n}{C}_3:{}^n{C}_2=12:1\left(ii\right){}^{2n}{C}_3:{}^n{C}_3=11:1$

Answer 1

(i) Here ${}^{2n}{C}_3{:}^n{C}_2=12:1$

$\Rightarrow \frac{\left(2n\right)!}{3!(2n-3)!}\times \frac{2!(n-2)!}{n!}=\frac{12}{1}$

$\Rightarrow \frac{\left(2n\right)(2n-1)(2n-2)(2n-3)!}{3\times 2!(2n-3)!}\times \frac{2!(n-2)!}{n(n-1)(n-2)!}=\frac{12}{1}$

$\Rightarrow \frac{\left(2n\right)(2n-1)(2n-2)}{3}\times \frac{1}{n(n-1)}=\frac{12}{1}$

$\Rightarrow \frac{4(2n-1)}{3}=\frac{12}{1}$

$\Rightarrow 8n-4=36\Rightarrow n=5$

(ii) Here ${}^{2n}{C}_3:{}^n{C}_3=11:1$

$\Rightarrow \frac{\left(2n\right)!}{3!(2n-3)!}\times \frac{2!(n-3)!}{n!}=\frac{11}{1}$

$\Rightarrow \frac{\left(2n\right)(2n-1)(2n-2)(2n-3)!}{3!(2n-3)!}\times \frac{3!(n-3)!}{n(n-1)(n-2)(n-3)!}=\frac{11}{1}$

$\Rightarrow \frac{\left(2n\right)(2n-1)(2n-2)}{n(n-1)(n-2)}=\frac{11}{1}$

$\Rightarrow \frac{4(2n-1)}{n-2}=\frac{11}{1}$

$\Rightarrow 8n-4=11n-22$

$\Rightarrow 3n=18\Rightarrow n=6.$