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Solution:

Given: ${ }^{2 n} C_{3}:{ }^{n} C_{3}=12: 1$

To find: $\mathrm{n}=$ ?

${ }^{2 n} \mathrm{C}_{3}:{ }^{n} \mathrm{C}_{3}=12: 1$

$\Rightarrow \frac{\left(\begin{array}{c}2 n \\ 3\end{array}\right)}{\left(\begin{array}{c}n \\ 3\end{array}\right)}=\frac{12}{1}$

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

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

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

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

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

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

$\Rightarrow 2 n-1=3(n-2)$

$\Rightarrow 2 n-1=3 n-6$

$\Rightarrow n=6-1=5$

Ans: $n=5$