The total number of natural numbers of six digits that can be made with digits $1, 2, 3, 4$,if all digits are to appear in the same number at least once,is

The units digit of $\sum\limits_{1 < p < 100} {p! - \sum\limits_{n = 1}^{50} {(2n)!} }$ is:

There are $3$ sections in a question paper and each section contains $5$ questions. $A$ candidate has to answer a total of $5$ questions,choosing at least one question from each section. Then the number of ways,in which the candidate can choose the questions,is

The number of seven-digit integers with the sum of the digits equal to $10$ and formed by using the digits $1, 2,$ and $3$ only is:

$5$ boys and $6$ girls are arranged in all possible ways. Let $X$ denote the number of linear arrangements in which no two boys sit together and $Y$ denote the number of linear arrangements in which no two girls sit together. If $Z$ denotes the number of ways of arranging all of them around a circular table such that no two boys sit together,then $X: Y: Z=$

$A$ set contains $2n + 1$ elements. The number of subsets of this set containing more than $n$ elements is equal to