QJEEM23Ma5b1d: JEE Main 2023 Mathematics - Class 11 JEEM23Ma5b1d Solution: Let A = { 1 , 3 , 4 , 6 , 9 } and B = { 2 , 4 , 5 , 8 , 10 } .…

Given that the sets are $A = \{1, 3, 4, 6, 9\}$ and $B = \{2, 4, 5, 8, 10\}$ , for the relation $\mathrm{R}$ on the set $A \times B$ , we need to find the combinations of pairs that satisfy the conditions $a_1 \leq b_2$ and $b_1 \leq a_2$ .

We find the number of combinations by considering the possible values for $b_2$ for each $a_1$ and the possible values for $a_2$ for each $b_1$ :

For each $a_1$ in $A = \{1, 3, 4, 6, 9\}$ , the number of valid $b_2$ values in $B = \{2, 4, 5, 8, 10\}$ are :

- For $a_1 = 1$ , there are 5 choices for $b_2$ .

- For $a_1 = 3$ , there are 4 choices for $b_2$ .

- For $a_1 = 4$ , there are 4 choices for $b_2$ .

- For $a_1 = 6$ , there are 2 choices for $b_2$ .

- For $a_1 = 9$ , there is 1 choice for $b_2$ .

This results in a total of $5+4+4+2+1 = 16$ possible pairs $(a_1, b_2)$ .

Similarly, for each $b_1$ in $B$ , the number of valid $a_2$ values in $A$ are :

- For $b_1 = 2$ , there are 4 choices for $a_2$ .

- For $b_1 = 4$ , there are 3 choices for $a_2$ .

- For $b_1 = 5$ , there are 2 choices for $a_2$ .

- For $b_1 = 8$ , there is 1 choice for $a_2$ .

- For $b_1 = 10$ , there are no choices for $a_2$ .

This results in a total of $4+3+2+1+0 = 10$ possible pairs $(b_1, a_2)$ .

Therefore, the total number of elements in the relation $\mathrm{R}$ , which satisfies the given conditions, is $16 \times 10 = 160$ .

So, the correct answer is 160.