JEE MAIN - Mathematics (2023 - 11th April Evening Shift - No. 6)
Let $$\mathrm{A}=\{1,3,4,6,9\}$$ and $$\mathrm{B}=\{2,4,5,8,10\}$$. Let $$\mathrm{R}$$ be a relation defined on $$\mathrm{A} \times \mathrm{B}$$ such that $$\mathrm{R}=\left\{\left(\left(a_{1}, b_{1}\right),\left(a_{2}, b_{2}\right)\right): a_{1} \leq b_{2}\right.$$ and $$\left.b_{1} \leq a_{2}\right\}$$. Then the number of elements in the set R is :
180
26
52
160
Explanation
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.
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.
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