$5$ 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 $X: Y: Z=$

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:

Let $S_1 = \{(i, j, k) : i, j, k \in \{1, 2, \ldots, 10\}\}$,$S_2 = \{(i, j) : 1 \leq i < j + 2 \leq 10, i, j \in \{1, 2, \ldots, 10\}\}$,$S_3 = \{(i, j, k, l) : 1 \leq i < j < k < l, i, j, k, l \in \{1, 2, \ldots, 10\}\}$,$S_4 = \{(i, j, k, l) : i, j, k \text{ and } l \text{ are distinct elements in } \{1, 2, \ldots, 10\}\}$. If the total number of elements in the set $S_r$ is $n_r$ for $r = 1, 2, 3, 4$,then which of the following statements is (are) $TRUE$?
$(A) n_1 = 1000$
$(B) n_2 = 44$
$(C) n_3 = 220$
$(D) \frac{n_4}{12} = 420$

The word $UNIVERSITY$ is arranged randomly. The probability that both $I$s do not come together is:

Total number of $3$ letter words that can be formed from the letters of the word $'SAHARANPUR'$ is equal to

For $n \geq 2$,let $S_n$ denote the set of all subsets of $\{1, 2, \ldots, n\}$ such that no two elements are consecutive. For example,$\{1, 3, 5\} \in S_6$,but $\{1, 2, 4\} \notin S_6$. Then $n(S_5)$ is equal to . . . . . . .