$3$ boys $B_i, i = 1, 2, 3$ and $6$ girls $G_i, i = 1, 2, . . . , 6$ are to be seated in a row. The number of ways they can be seated so that $B_1, B_2$ are separated and $G_1, G_2$ are also separated is equal to:

For a natural number $n$,the inequality $2^n(n - 1)! < n^n$ holds true if:

Five boys and five girls are writing an examination in a hall in which $5$ benches are arranged in a row and only two students are to be seated on every bench at either of its ends. If the seating arrangement is to be such that no two boys or no two girls sit together as neighbours (a student should not have a student of same gender either on left or right),then the total number of such arrangements is

If $f(n) = n! (31-n)!$,where $n \in \{0, 1, 2, \ldots, 31\}$,then the minimum value of $f(n)$ is

There are $3$ applicants for a scholarship in Physics,$5$ for Mathematics,and $4$ for Chemistry. In how many different ways can one of these scholarships be awarded?

The number of ways to distribute $30$ identical candies among four children $C_{1}, C_{2}, C_{3}$ and $C_{4}$ such that $C_{2}$ receives at least $4$ and at most $7$ candies,and $C_{3}$ receives at least $2$ and at most $6$ candies,is equal to