Question

If ${}^{2n}{C}_3:$ ${}^n{C}_3=11:1$, then $n=$ ________.

Answer 1

Given:

${}^{2n}{C}_3{:}^n{C}_3=11:1$

${}^{2n}{C}_3=\frac{2n!}{3!\left(2n-3\right)!}$

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

$=\frac{\left(2n\right)\left(2n-1\right)\left(2n-2\right)}{6}$

${}^n{C}_3=\frac{n!}{3!\left(n-3\right)!}$

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

$=\frac{\left(n\right)\left(n-1\right)\left(n-2\right)}{6}$

The expression becomes,

$\frac{\left(2n\right)\left(2n-1\right)\left(2n-2\right)}{6}:\frac{\left(n\right)\left(n-1\right)\left(n-2\right)}{6}=11:1$

$\frac{\frac{\left(2n\right)\left(2n-1\right)\left(2n-2\right)}{6}}{\frac{\left(n\right)\left(n-1\right)\left(n-2\right)}{6}}=\frac{11}{1}$

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

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

$4\left(2n-1\right)=11\left(n-2\right)$

$8n-4=11n-22$

$-4+22=11n-8n$

$n=6$

Hence, the value of n is 6.