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Let $A=\{1,3,4,6,9\}$ and $B=\{2,4,5,8,10\}$. Let $R$ be a relation defined on $A \times B$ such that $R =$ $\left\{\left(\left(a_1, b_1\right),\left(a_2, b_2\right)\right): a_1 \leq b_2\right.$ and $\left.b_1 \leq a_2\right\}$. Then the number of elements in the set $R$ is
$26$
$160$
$180$
$52$
Solution
Let $a_1=1 \Rightarrow 5$ choices of $b _2$
$a_1=3 \Rightarrow 4 \text { choices of } b_2$
$a_1=4 \Rightarrow 4 \text { choices of } b_2$
$a_1=6 \Rightarrow 2 \text { choices of } b_2$
$a_1=9 \Rightarrow 1 \text { choices of } b_2$
For $\left(a_1, b _2\right) 16$ ways .
Similarly, $b_1=2 \Rightarrow 4$ choices of $a_2$
$b _1=4 \Rightarrow 3 \text { choices of } a _2$
$b _1=5 \Rightarrow 2 \text { choices of } a _2$
$b _1=8 \Rightarrow 1 \text { choices of } a _2$
Required elements in $R =160$
