If $^{n}P_{4} = 24 \times ^{n}C_{5}$,then the value of $n$ is

Let $n = 1! + 4! + 7! + \ldots + 400!$. Then the ten's digit of $n$ is

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

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 . . . . . . . .

The value of $\sum_{r=0}^{20} {}^{50-r}C_{6}$ is equal to

$A$ $5$-digit number $\overline{abcde}$,when multiplied by $9$,gives the $5$-digit number $\overline{edcba}$. The sum of the digits in the number is: