Out of $800$ boys in a school,$224$ played cricket,$240$ played hockey,and $336$ played basketball. Of the total,$64$ played both basketball and hockey,$80$ played cricket and basketball,and $40$ played cricket and hockey; $24$ played all three games. The number of boys who did not play any game is:

In a class of $140$ students numbered $1$ to $140$,all even-numbered students opted for the Mathematics course,those whose number is divisible by $3$ opted for the Physics course,and those whose number is divisible by $5$ opted for the Chemistry course. The number of students who did not opt for any of the three courses is:

If ${N_a} = \{an : n \in N\}$,then ${N_5} \cap {N_7} = $

Let $A = \{a, b, c\}$ and $B = \{1, 2, 3, 4\}$. Then the number of elements in the set $C = \{f : A \rightarrow B \mid 2 \in f(A) \text{ and } f \text{ is not one-one}\}$ is

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

Suppose $A_1, A_2, A_3, \dots, A_{30}$ are $30$ sets each having $5$ elements and $B_1, B_2, \dots, B_n$ are $n$ sets each with $3$ elements. Let $\bigcup_{i=1}^{30} A_i = \bigcup_{j=1}^n B_j = S$ and each element of $S$ belongs to exactly $10$ of the $A_i$'s and exactly $9$ of the $B_j$'s. Then $n$ is equal to: