Determine $n$ if
(i) ${}^{2n}{C}_3:{}^n{C}_3=12:1$
(ii) ${}^{2n}{C}_3:{}^n{C}_3=11:1$

Solution: (i) $C\left(2n,2\right):C\left(n,2\right)=\frac{\frac{2n!}{\left(2!\right)\left(2n-2\right)!}}{\frac{n!}{\left(2!\right)\left(n-2\right)!}}$
$=\frac{\frac{\left(2n\right)\ast \left(2n-1\right)\ast \left(2n-2\right)!}{\left(2!\right)\left(2n-2\right)!}}{\frac{\left(n\right)\ast \left(n-1\right)\ast \left(n-2\right)!}{\left(2!\right)\left(n-2\right)!}}$
$=\frac{2\left(2n-1\right)}{n-1}$
Now, we are given,$C\left(2n,2\right):C\left(n,2\right)=12:1$
$\therefore \frac{2\left(2n-1\right)}{n-1}=12$
$\Rightarrow 4n-2=12n-12\Rightarrow 8n=10\Rightarrow n=\frac{5}{4}$
But, $n$ should be a whole number.
So, there is no value of $n$ that can solve this.

(ii) $C\left(2n,3\right):C\left(n,3\right)=\frac{\frac{2n!}{\left(3!\right)\left(2n-3\right)!}}{\frac{n!}{\left(3!\right)\left(n-3\right)!}}$
$=\frac{\frac{\left(2n\right)\ast \left(2n-1\right)\ast \left(2n-2\right)\ast \left(2n-3\right)!}{\left(2!\right)\left(2n-3\right)!}}{\frac{\left(n\right)\ast \left(n-1\right)\ast \left(n-2\right)\ast \left(n-3\right)!}{\left(3!\right)\left(n-3\right)!}}$
$=\frac{2\left(2n-1\right)\left(2n-2\right)}{\left(n-1\right)\left(n-2\right)}$
$=\frac{4\left(2n-1\right)}{n-2}$
Now, we are given,$C\left(2n,3\right):C\left(n,3\right)=11:1$
$\therefore \frac{4\left(2n-1\right)}{n-2}=\frac{11}{1}$
$\Rightarrow 8n-4=11n-22$
$\Rightarrow 3n=18\Rightarrow n=6$