The number of integers greater than $6000$ that can be formed using the digits $3, 5, 6, 7,$ and $8$ without repetition is:

If the number of circular permutations of $9$ distinct things taken $5$ at a time is $n_1$ and the number of linear permutations of $8$ distinct things taken $4$ at a time is $n_2$,then $\frac{n_1}{n_2}=$

$5$ 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 $X: Y: Z=$

Consider the following two statements:
$I.$ If $n$ is a composite number,then $n$ divides $(n-1)!$.
$II.$ There are infinitely many natural numbers $n$ such that $n^3+2n^2+n$ divides $n!$.

There are three sections in a question paper,each containing $4$ questions. If a candidate has to answer only $5$ questions from this paper without leaving any section,then the number of ways the candidate can make the choice of questions is:

The number of all $3$-digit numbers $abc$ (in base $10$) for which $(a \times b \times c) + (a \times b) + (b \times c) + (c \times a) + a + b + c = 29$ is: