If  'R' is a relation from 'A' to 'B' , then 'R' is defined as {(x,y)| x$\in$ A and y$\in$ B}
In all the ordered pairs in the realtions of R${}_1$, R${}_2$, R${}_3$ first component $\in$ X and second component $\in$ Y.  So, R${}_1$, R${}_2$, R${}_3$are relations from X to Y.
$R_4$ is not a relation from X to Y , because in ordered pain (7,9) the first component 7 $\notin$ X.

Since A $\subseteq$ A . $\therefore$ Relation ' $\subseteq$ ' is relfexive 

Since A $\subseteq$ B , B $\subseteq$ C $\Rightarrow$ A $\subseteq$ C 

$\therefore$ Relation ' $\subseteq$ ' is transitive. 

But A ' $\subseteq$ ' B , $\Rightarrow$ B  $\subseteq$ A . $\therefore$ relation is not symmetric. 

We have, $R$={(1,3);(1,5);(2,3);(2,5);(3,5);(4,5)}
$R^{-1}$ = {(3,1);(5,1);(3,2);(5,2);(5,3);(5,4)}
Hence $Ro{R}^{-1}$ = {(3,3);(3,5);(5,3);(5,5)}

Here n(A × B) = 3 × 3 = 9 

Since every subset of A × B defines a relation from A to B, the number of relations from A to B is equal to the number of subsets of A $\times$ B = $2^{\text{n(A}\times \text{B)}}$
                                               = $2^9$
                                               = 512

n ( A × A) = n(A)n(A) = $3^2$ = 9 

So, the total number of subsets of A × A is $2^9$ 

and a subset of A × A is a relation over the set A . 

$R=\left\{\right(a,b):a,b\in N,a-b=3\}=\left\{\right((n+3),n):n\in N\}$ 

={(4,1),(5,2),(6,3), .......} . 

R = {(2,1),(4,2),(6,3),.....}.

So, $R^{-1}$ = {(1,2),(2,4),(3,6),......}.

Given, A = {$x:x^2-5x+6=0$}
$\therefore$ The elements of A are the roots of $x^2-5x+6=0$
 $x^2-5x+6=0\Rightarrow (x-3)(x-2)=0\Rightarrow x=3\text{and}2$
$\therefore$ A = {2,3} , B = {2,4}, C = {4,5}

$\Rightarrow B\cap C$ = {4}

$\therefore$ A × (B $\cap$ C) = {2, 3} $\times$ {4}
                        ={(2,4), (3,4)}

( x,y) $\in$ R ⇔ (y,x) $\in R^{-1},$       $\therefore$ $R^{-1}=\left\{\right(3,1),(5,1),(1,2\left)\right\}$.

A relation from P to Q is a subset of P $\times$ Q.