Question

If ${}_{}^{2n}{C}_3^{}:{}_{}^n{C}_3^{}=11:1$ , then find the value of $n.$

Answer 1

Answer: 6

It is given that, ${}_{}^{2n}{C}_3^{}:{}_{}^n{C}_3^{}=11:1$
$\text{⇒}\frac{2n\left(2n-1\right)\left(2n-2\right)}{\mathrm{1.2.3}}÷\frac{n\left(n-1\right)\left(n-2\right)}{\mathrm{1.2.3}}=\frac{11}{1}$
$\text{⇒}\frac{4n\left(n-1\right)\left(2n-1\right)}{6}\times \frac{6}{n\left(n-1\right)\left(n-2\right)}=\frac{11}{1}$
$\Rightarrow \frac{4\left(2n-1\right)}{n-2}=11$
$\Rightarrow 4(2n-1)=11(n-2)$
$\Rightarrow 8n-4=11n-22$
$\Rightarrow 3n=18$
$\Rightarrow n=6$