JEE Main 2023 — Sets And Relations Question with Solution

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 $a_1\le b_2$ and $b_1\le a_2$.

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.