JEE Mathematics 2023 Question with Solution

Given: $A = \{1, 3, 4, 6, 9\}$, $B = \{2, 4, 5, 8, 10\}$, and $R = \{((a_1,b_1),(a_2,b_2)) : a_1 \leq b_2 \text{ and } b_1 \leq a_2\}$.

Find: The number of elements in $R$.

Count the valid pairs for the condition $a_1 \leq b_2$.

For $a_1 = 1$, there are $5$ choices of $b_2$. For $a_1 = 3$, there are $4$ choices of $b_2$. For $a_1 = 4$, there are $4$ choices of $b_2$. For $a_1 = 6$, there are $2$ choices of $b_2$. For $a_1 = 9$, there is $1$ choice of $b_2$.

So the total number of choices for $(a_1,b_2)$ is

$$5 + 4 + 4 + 2 + 1 = 16$$

Now count the valid pairs for the condition $b_1 \leq a_2$.

For $b_1 = 2$, there are $4$ choices of $a_2$. For $b_1 = 4$, there are $3$ choices of $a_2$. For $b_1 = 5$, there are $2$ choices of $a_2$. For $b_1 = 8$, there is $1$ choice of $a_2$. For $b_1 = 10$, there are $0$ choices of $a_2$.

So the total number of choices for $(b_1,a_2)$ is

$$4 + 3 + 2 + 1 + 0 = 10$$

Therefore, the total number of elements in $R$ is

$$16 \times 10 = 160$$

The required number of elements is $160$.

Although the solution states "The Correct Option is C", the computed value $160$ matches option B among the given options. Hence, the correct option is B.

Given: The relation condition splits into two independent inequalities: $a_1 \leq b_2$ and $b_1 \leq a_2$.

Find: Use separate counting and multiply the results.

The first inequality involves only $a_1$ and $b_2$, while the second involves only $b_1$ and $a_2$. Therefore, the total count is the product of the number of valid choices for these two parts.

1. Count valid $(a_1,b_2)$ pairs.
2. Count valid $(b_1,a_2)$ pairs.
3. Multiply the two totals.

This gives

$$|R| = 16 \times 10 = 160$$

So the correct option is B.