The number of ways in which $3$ children can distribute $10$ tickets out of $15$ consecutively numbered tickets such that they get consecutive blocks of $5$,$3$,and $2$ tickets is:

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 . . . . . . . .

How many six-digit numbers are there in which no digit is repeated,even digits appear at even places,odd digits appear at odd places,and the number is divisible by $4$?

If $\sum\limits_{r = 0}^{25} {\left( {^{50}C_r \cdot ^{50 - r}C_{25 - r}} \right) = K\left( {^{50}C_{25}} \right)}$,then $K$ is equal to

The sum of all $4$-digit numbers that can be formed using the digits $2, 3, 4, 5, 6$ without repetition is:

The value of $(2 \cdot {}^{1}P_{0} - 3 \cdot {}^{2}P_{1} + 4 \cdot {}^{3}P_{2} - \dots \text{ up to } 51^{\text{th}} \text{ term}) + (1! - 2! + 3! - \dots \text{ up to } 51^{\text{th}} \text{ term})$ is equal to