Let $A$ be a set containing $n$ elements. If $P$ and $Q$ are two subsets of $A$, then the number of ways of choosing $P$ and $Q$ such that $P \cap Q = \phi$ is:

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:

The solution set of the system of equations $x + y = \frac{2\pi}{3}$ and $\cos x + \cos y = \frac{3}{2}$,where $x$ and $y$ are real,is:

In a college of $300$ students,every student reads $5$ newspapers and every newspaper is read by $60$ students. The number of newspapers is

Sets $A$ and $B$ have $3$ and $6$ elements respectively. What can be the minimum number of elements in $A \cup B$?

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