Let A = \ 1, 3, 4, 6, 9\ and B = \ 2, 4, 5, 8, 10\ . Let R be a...

Given,

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

And relation is given by,

R = \left\{\left({a}_{1}, {b}_{1}), \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 $= 16 \times 10 = 160$