The number of integers $n$ such that $100 \leq n \leq 999$ and $n$ contains at most two distinct digits is:

The remainder obtained when $1! + 2! + 3! + \dots + 11!$ is divided by $12$ is

$A$ five-digit number divisible by $3$ has to be formed using the numerals $0, 1, 2, 3, 4,$ and $5$ without repetition. The total number of ways in which this can be done is:

$A$ building has a ground floor and $10$ more floors. Nine persons enter a lift at the ground floor. The lift goes up to the $10$th floor. The number of ways in which any $4$ persons exit at a floor and the remaining $5$ persons exit at a different floor, if the lift does not stop at the first and the second floors, is equal to:

The number $(49^{2}-4)(49^{3}-49)$ is divisible by (in $!$)

In how many ways can $5$ distinct balls be distributed among $3$ persons such that each person receives at least one ball?