Let A = {1, 3, 4, 6, 9} and B = {2, 4, 5, 8, 10}. Let R be a relation defined on A x B such that and Then the number of elements in the set R is [2023]

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Solution

Q.
Let A = {1, 3, 4, 6, 9} and B = {2, 4, 5, 8, 10}. Let R be a relation defined on A x B such that $R = {((a_{1}, b_{1}), (a_{2}, b_{2})) : a_{1} \leq b_{2}$ and $b_{1} \leq a_{2}} .$ Then the number of elements in the set R is [2023]

1 52

2 180

3 26

4 160

Ans.
(4)

We have, $A = {1, 3, 4, 6, 9}$ ; $B = {2, 4, 5, 8, 10}$

$R = {((a_{1}, b_{1}), (a_{2}, b_{2})) : a_{1} \leq b_{2} and b_{1} \leq a_{2}}$

At $a_{1} = 1$ , there are 5 choices for $b_{2}$ ;

$a_{1} = 3$ , there are 4 choices for $b_{2}$ ;

$a_{1} = 4$ , there are 4 choices for $b_{2}$ ;

$a_{1} = 6$ , there are 2 choices for $b_{2}$ ;

$a_{1} = 9$ , there is 1 choice for $b_{2}$ ;

$∴$ $Total ways for (a_{1} \leq b_{2}) = 16$

Now, at $b_{1} = 2$ , there are 4 choices for $a_{2}$ ;

$b_{1} = 4$ , there are 3 choices for $a_{2}$ ;

$b_{1} = 5$ , there are 2 choices for $a_{2}$ ;

$b_{1} = 8$ , there is 1 choice for $a_{2}$ .

$Total ways for (b_{1} \leq a_{2}) = 10$

$∴ Required number of ways = 16 \times 10 = 160$

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