JEE MAIN - Mathematics (2023 - 10th April Evening Shift - No. 12)
Let $$\mathrm{A}=\{2,3,4\}$$ and $$\mathrm{B}=\{8,9,12\}$$. Then the number of elements in the relation
$$\mathrm{R}=\left\{\left(\left(a_{1}, \mathrm{~b}_{1}\right),\left(a_{2}, \mathrm{~b}_{2}\right)\right) \in(A \times B, A \times B): a_{1}\right.$$ divides $$\mathrm{b}_{2}$$ and $$\mathrm{a}_{2}$$ divides $$\left.\mathrm{b}_{1}\right\}$$ is :
18
24
36
12
Explanation
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 \times 6 = 36 $
relations.
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