If $A$ is the set of even natural numbers less than $8$ and $B$ is the set of prime numbers less than $7$, then the number of relations from $A$ to $B$ is

Let the relations $R_1$ and $R_2$ on the set $\mathrm{X}=\{1,2,3, \ldots, 20\}$ be given by $\mathrm{R}_1=\{(\mathrm{x}, \mathrm{y}): 2 \mathrm{x}-3 \mathrm{y}=2\}$ and $\mathrm{R}_2=\{(\mathrm{x}, \mathrm{y}):-5 \mathrm{x}+4 \mathrm{y}=0\}$. If $\mathrm{M}$ and $\mathrm{N}$ be the minimum number of elements required to be added in $R_1$ and $R_2$, respectively, in order to make the relations symmetric, then $\mathrm{M}+\mathrm{N}$ equals

Show that the relation $\mathrm{R}$ in the set $\mathrm{Z}$ of integers given by $\mathrm{R} =\{(\mathrm{a}, \mathrm{b}): 2$ divides $\mathrm{a}-\mathrm{b}\}$ is an equivalence relation.

Let a relation $R$ be defined by $R = \{(4, 5); (1, 4); (4, 6); (7, 6); (3, 7)\}$ then ${R^{ - 1}}oR$ is

A relation on the set $A\, = \,\{ x\,:\,\left| x \right|\, < \,3,\,x\, \in Z\} ,$ where $Z$ is the set of integers is defined by $R= \{(x, y) : y = \left| x \right|, x \ne  - 1\}$. Then the number of elements in the power set of $R$ is

Let $A=\{1,2,3\} .$ Then number of equivalence relations containing $(1,2)$ is