A gaurd of $12$ men is formed from a group of`n` soliders. It is found that $2$ particular soliders A and B are $3$ times as often together on guard as $3$ particulars soliders C,D and E. Then n is equal to

Total number of men$=12$

Group of soldiers $=n$

First we found that $2$ particular soldier$=12-2=10$

Then, the value is  ${}^{n-2}{C}_{10}$    ………     (1)

Now we found that 3 times as 3 particulars soldiers $=12-3=9$

Then, the value is    $3{\times }^{n-3}{C}_9$    ………    (2)

According to given question

By equation (1) and (2) to and we get,

${}^{n-2}{C}_{10}=3{\times }^{n-3}{C}_9$

$\Rightarrow \frac{\left(n-2\right)!}{10!\left(n-2-10\right)!}=3\times \frac{\left(n-3\right)!}{9!\left(n-3-9\right)!}$

$\Rightarrow \frac{\left(n-2\right)\left(n-3\right)!}{10\times 9!\left(n-12\right)!}=3\times \frac{\left(n-3\right)!}{9!\left(n-12\right)!}$

$\Rightarrow \frac{\left(n-2\right)}{10}=3$

$\Rightarrow n-2=30$

$\Rightarrow n=32$

Hence, it is complete solution .

Option $C$  is correct.