Let A=2,3,4 and B=8,9,12.

Question

Let $A={2,3,4}$ and $B={8,9,12}$. Then the number of elements in the relation

$R={(({a}_{1},{b}_{1}),({a}_{2},{b}_{2}))\in (A	imes B,A	imes B):{a}_{1}$ divides ${b}_{2}$ and ${a}_{2}$ divides ${b}_{1}}$ is :

Accepted Answer

Given sets :

$A=2,3,4$

$B=8,9,12$

We want to find the number of elements of the form $(({a}_{1},{b}_{1}),({a}_{2},{b}_{2}))$ such that :

${a}_{1}$ divides ${b}_{2}$
${a}_{2}$ divides ${b}_{1}$

For the first condition :

${a}_{1}$ divides ${b}_{2}$

Given ${a}_{1}\in A$ and ${b}_{2}\in B$, we can list the pairs:

$({a}_{1},{b}_{2})\in (2,8),(2,12),(3,9),(3,12),(4,8),(4,12)$

This gives 6 pairs.

For the second condition, the pairs are the same, because it's just the reversed relation. So :

${a}_{2}$ divides ${b}_{1}$

Again has 6 valid pairs.

Now, for every pair from the first condition, we can have any pair from the second condition. This leads to :

$6	imes 6=36$
relations.

Solutions · 1