The remainder obtained when $1! + 2! + 95!$ is divided by $15$ is

$(2n + 1) (2n + 3) (2n + 5) \dots (4n - 1)$ is equal to :

Ten students are seated at random in a row. The probability that two particular students are not seated side by side is

$A$ man $X$ has $7$ friends,$4$ of them are ladies and $3$ are men. His wife $Y$ also has $7$ friends,$3$ of them are ladies and $4$ are men. Assume $X$ and $Y$ have no common friends. The total number of ways in which $X$ and $Y$ together can throw a party inviting $3$ ladies and $3$ men,such that $3$ friends of each of $X$ and $Y$ are invited,is:

Let $m$ be a natural number such that $20000 < m < 60000$ and let $k$ be the sum of all the digits in $m$. Then the number of numbers $m$ for which $k$ is even,is

In how many ways can the letters of the word $ASSASSIN$ be arranged so that no two $S$s occur together?