The remainder obtained when $1! + 2! + 3! + \dots + 11!$ is divided by $12$ is

The number $(49^{2}-4)(49^{3}-49)$ is divisible by (in $!$)

Let $x$ denote the number of ways of arranging $m$ boys and $m$ girls in a row so that no two boys sit together. If $y$ and $z$ denote the number of ways of arranging $m$ boys and $m$ girls in a row and around a circular table respectively so that boys and girls sit alternately,then $x: y: z=$

Let $S$ be the set of all permutations $a_1, a_2, \ldots, a_6$ of $1, 2, \ldots, 6$ such that $a_1, a_2, \ldots, a_k$ is not a permutation of $1, 2, \ldots, k$ for any $k, 1 \leq k \leq 5$. Then the number of elements in $S$ is:

$A$ man $P$ has $7$ friends,$4$ of them are ladies and $3$ are men. His wife $Q$ also has $7$ friends,$3$ of them are ladies and $4$ are men. Assume $P$ and $Q$ have no common friends. Then the total number of ways in which $P$ and $Q$ together can throw a party inviting $3$ ladies and $3$ men,so that $3$ friends of each of $P$ and $Q$ are in this party,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: