Among the $4$-digit numbers that can be formed using the digits $1, 2, 3, 4, 5, 6$ without repeating any digit,the number of numbers which are divisible by $6$ is:

$A$ question paper has $3$ parts $A, B, C$. Part $A$ contains $7$ questions,part $B$ contains $5$ questions,and part $C$ contains $3$ questions. If a candidate is allowed to answer not more than $4$ questions from part $A$,not more than $3$ questions from part $B$,and not more than $2$ questions from part $C$,then the number of ways in which a candidate can answer exactly $7$ questions is

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

Numbers between $1$ and $10,000$ are formed using the digits $2$ and $3$ exactly once and the digit $4$ twice. If the numbers thus formed are arranged in increasing order and $x, y$ represent the ranks of $4324$ and $324$ respectively,then $x-y=$

If $\frac{1}{8!} + \frac{1}{9!} = \frac{x}{10!},$ find $x.$

Find the remainder when $(1!)^2 + (2!)^2 + (3!)^2 + \dots + (100!)^2$ is divided by $10^2$.