Let $a>0, a \neq 1$. Then, the set $S$ of all positive real numbers $b$ satisfying $\left(1+a^2\right)\left(1+b^2\right)=4 a b$ is

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

Let $S=\{1,2,3, \ldots \ldots, n\}$ and $A=\{(a, b) \mid 1 \leq$ $a, b \leq n\}=S \times S$. A subset $B$ of $A$ is said to be a good subset if $(x, x) \in B$ for every $x \in S$. Then, the number of good subsets of $A$ is

Let $\mathrm{A}=\{\mathrm{n} \in[100,700] \cap \mathrm{N}: \mathrm{n}$ is neither a multiple of $3$ nor a multiple of 4$\}$. Then the number of elements in $\mathrm{A}$ is

If $X = \{ {8^n} - 7n - 1:n \in N\} $ and $Y = \{ 49(n - 1):n \in N\} ,$ then

The number of elements in the set $\left\{ n \in N : 10 \leq n \leq 100\right.$ and $3^{ n }-3$ is a multiple of $7\}$ is $........$.