Determine $n$ if
$^{2 n} C_{3}:\,^{n} C_{3}=12: 1$
$\frac{{^{2n}{C_3}}}{{^n{C_3}}} = \frac{{12}}{1}$
$\Rightarrow \frac{(2 n) !}{3 !(2 n-3) !} \times \frac{3 !(n-3) !}{n !}=\frac{12}{1}$
$\Rightarrow \frac{(2 n)(2 n-1)(2 n-2)(2 n-3) !}{(2 n-3) !} \times \frac{(n-3) !}{n(n-1)(n-2)(n-3) !}=12$
$\Rightarrow \frac{2(2 n-1)(2 n-2)}{(n-1)(n-2)}=12$
$\Rightarrow \frac{4(2 n-1)(n-1)}{(n-1)(n-2)}=12$
$\Rightarrow \frac{(2 n-1)}{(n-2)}=3$
$\Rightarrow 2 n-1=3(n-2)$
$\Rightarrow 2 n-1=3 n-6$
$\Rightarrow 3 n-2 n=-1+6$
$\Rightarrow n=5$
The number of ways in which an examiner can assign $30$ marks to $8$ questions, giving not less than $2$ marks to any question, is
From $6$ different novels and $3$ different dictionaries, $4$ novels and $1$ dictionary are to be selected and arranged in a row on a shelf so that the dictionary is always in the middle. The number of such arrangements is :
$^{47}{C_4} + \mathop \sum \limits_{r = 1}^5 {}^{52 - r}{C_3} = $
A group consists of $4$ girls and $7$ boys. In how many ways can a team of $5$ members be selected if the team has no girl?
A car will hold $2$ in the front seat and $1$ in the rear seat. If among $6$ persons $2$ can drive, then the number of ways in which the car can be filled is