In a hotel,four rooms are available. Six persons are to be accommodated in these four rooms in such a way that each of these rooms contains at least one person and at most two persons. Then the number of all possible ways in which this can be done 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:

If $n = ^mC_2$,then the value of $^nC_2$ is given by

$A$ bag contains $n$ white and $n$ black balls. Pairs of balls are drawn at random without replacement successively,until the bag is empty. If the number of ways in which each pair consists of one white and one black ball is $14400$,then $n$ is equal to

Let $S$ be the set of all passwords which are $6$ to $8$ characters long,where each character is either an alphabet from $\{A, B, C, D, E\}$ or a number from $\{1, 2, 3, 4, 5\}$ with the repetition of characters allowed. If the number of passwords in $S$ whose at least one character is a number from $\{1, 2, 3, 4, 5\}$ is $\alpha \times 5^{6}$,then $\alpha$ is equal to $.......$

The sum of all the $4$-digit numbers formed by taking all the digits from ${2, 3, 5, 7}$ without repetition is: