If $U = \{a, b, c, d, e, f\}$,$A = \{a, b, c\}$,$B = \{c, d, e, f\}$,and $C = \{c, d, e\}$,find $(A \cap B) \cup (A \cap C)$.

Let $S = \{ x \in R : x \ge 0 \text{ and } 2|\sqrt{x} - 3| + \sqrt{x}(\sqrt{x} - 6) + 6 = 0 \}$. Then $S$:

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:

If $A = \{a, d\}, B = \{b, c, e\}$ and $C = \{b, c, f\},$ then $A \times (B \cap C) =$

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

If $n(A) = 4$,$n(B) = 3$,and $n(A \times B \times C) = 24$,then $n(C) = $