From a collection of $20$ consecutive natural numbers, four are selected such that they are not consecutive. The number of such selections is

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:

All possible words (with or without meaning) are formed by taking at least $2$ letters (all different) from the letters of the word '$CURVE$'. If a word is chosen at random from all the words thus formed,then the probability of getting a $3$-letter word is

$3$ boys $B_i, i = 1, 2, 3$ and $6$ girls $G_i, i = 1, 2, . . . , 6$ are to be seated in a row. The number of ways they can be seated so that $B_1, B_2$ are separated and $G_1, G_2$ are also separated is equal to:

Number of all possible words (with or without meaning) that can be formed using all the letters of the word $CABINET$ in which neither the word $CAB$ nor the word $NET$ appear 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 $X: Y: Z=$