Among the $4$-digit numbers that can be formed using the digits $1, 2, 3, 4, 5, 6$ without repeating any digit,the number of numbers which are divisible by $6$ is:

Let $a_1, a_2, \ldots, a_n$ be $n$ non-zero real numbers,of which $p$ are positive and the remaining are negative. The number of ordered pairs $(j, k)$ with $j < k$ for which $a_j a_k$ is positive is $55$. Similarly,the number of ordered pairs $(j, k)$ with $j < k$ for which $a_j a_k$ is negative is $50$. Then,the value of $p^2 + (n-p)^2$ is

What is the remainder when $1! + 2! + 3! + \dots + 200!$ is divided by $14$?

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 students $S_{1}, S_{2}, \ldots, S_{10}$ are to be divided into $3$ groups $A, B$ and $C$ such that each group has at least one student and the group $C$ has at most $3$ students. Then the total number of possibilities of forming such groups is ........ .

The total number of three-digit and five-digit integers which can be formed by using the digits $0, 1, 2, 3, 4, 5$,using each digit not more than once in each number,is: