Let $A=\left\{1,3,4,6,9\right\}$ and $B=\left\{2,4,5,8,10\right\}$. Let $R$ be a relation defined on $A\times B$ such that  $R=\left\{\left(a_1,\left.b_1\right),\left(a_2,b_2\right):a_1\le b_2\right.\text{ and} b_1\le a_2\right\}$. Then the number of elements in the set $R$ is

Given,

$A=\left\{1,3,4,6,9\right\} \& B=\left\{2,4,5,8,10\right\}$

And relation is given by,

$R=\left\{\left(a_1,\left.b_1\right),\left(a_2,b_2\right):a_1\le b_2\right.\text{ and} b_1\le a_2\right\}$

Now taking cases for $a_1\le b_2$ we get,

$a_1=1, b_2\in \left\{2,4,5,8,10\right\}\to 5 \text{cases}$

$a_1=3, b_2\in \left\{4,5,8,10\right\}\to 4 \text{cases}$

$a_1=4, b_2\in \left\{4,5,8,10\right\}\to 4 \text{cases}$

$a_1=6, b_2\in \left\{8,10\right\}\to 2 \text{cases}$

$a_1=9, b_2\in \left\{10\right\}\to 1 \text{cases}$

So, total $16$ cases will be there 

Now finding cases of $b_1\le a_2$ we get,

$b_1=2, a_2\in \left\{3,4,6,9\right\}\to 4 \text{cases}$

$b_1=4, a_2\in \left\{4,6,9\right\}\to 3 \text{cases}$

$b_1=5, a_2\in \left\{6,9\right\}\to 2 \text{cases}$

$b_1=8, a_2\in \left\{9\right\}\to 1 \text{cases}$

So, here total $10$ cases,

Hence, total elements in relation $=16\times 10=160$