$2n(A \setminus B) = n(B \setminus A)$ and $5n(A \cap B) = n(A) + 3n(B)$,where $P \setminus Q = P \cap Q^C$. If $n(A \cup B) \leq 10$,then the value of $\frac{n(A) \cdot n(B) \cdot n(A \cap B)}{8}$ is:

The set $\{x \in R : [x - |x|] = 5\}$ is equal to

Let $A = \{1, 6, 11, 16, \dots\}$ and $B = \{9, 16, 23, 30, \dots\}$ be the sets consisting of the first $2025$ terms of two arithmetic progressions. Then $n(A \cup B)$ is

If $n(A) = 3$,$n(B) = 6$ and $A \subseteq B$. Then the number of elements in $A \cup B$ is equal to

In a survey,it was found that $21$ people liked product $A$,$26$ liked product $B$,and $29$ liked product $C$. If $14$ people liked products $A$ and $B$,$12$ people liked products $C$ and $A$,$14$ people liked products $B$ and $C$,and $8$ liked all three products,find how many people liked product $C$ only.

Of the members of three athletic teams in a school,$21$ are in the cricket team,$26$ are in the hockey team,and $29$ are in the football team. Among them,$14$ play hockey and cricket,$15$ play hockey and football,and $12$ play football and cricket. Eight play all the three games. The total number of members in the three athletic teams is: