Two finite sets $A$ and $B$ are such that $n(A) = 2$ and $n(B) = 3$. Then the total number of relations from $A$ to $B$ is:

$A$ relation $R$ is defined from $\{2, 3, 4, 5\}$ to $\{3, 6, 7, 10\}$ by $xRy \iff x$ is relatively prime to $y$. Then the domain of $R$ is

Write the relation $R = \{ (x, x^3) : x \text{ is a prime number less than } 10 \}$ in roster form.

$A = \{1, 2, 3, 5\}$ and $B = \{4, 6, 9\}$. Define a relation $R$ from $A$ to $B$ by $R = \{(x, y) : \text{the difference between } x \text{ and } y \text{ is odd}; x \in A, y \in B\}$. Write $R$ in roster form.

Let the relation $R$ be defined on the set of natural numbers $N$ by $a R b$ if $3 a+2 b=27$. Then $R$ is:

Let $A = \{1, 2, 3, 4, \ldots, 10\}$ and $B = \{0, 1, 2, 3, 4\}$. The number of elements in the relation $R = \{(a, b) \in A \times A : 2(a - b)^2 + 3(a - b) \in B\}$ is $.........$.