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JEE Main 2022
Sets, Relations & Functions
Sets and Relations
Medium

Question

Let A={2,3,4}\mathrm{A}=\{2,3,4\} and B={8,9,12}\mathrm{B}=\{8,9,12\}. Then the number of elements in the relation R={((a1, b1),(a2, b2))(A×B,A×B):a1\mathrm{R}=\left\{\left(\left(a_{1}, \mathrm{~b}_{1}\right),\left(a_{2}, \mathrm{~b}_{2}\right)\right) \in(A \times B, A \times B): a_{1}\right. divides b2\mathrm{b}_{2} and a2\mathrm{a}_{2} divides b1}\left.\mathrm{b}_{1}\right\} is :

Options

Solution

1. Key Concepts and Formulas

  • Cartesian Product: For sets XX and YY, the Cartesian product X×YX \times Y is the set of all ordered pairs (x,y)(x, y) where xXx \in X and yYy \in Y.
  • Divisibility: An integer aa divides an integer bb (denoted aba|b) if there exists an integer kk such that b=akb = ak.
  • Multiplication Principle of Counting: If there are n1n_1 ways to perform the first task and n2n_2 ways to perform the second task, and these tasks are independent, then there are n1×n2n_1 \times n_2 ways to perform both tasks.

2. Step-by-Step Solution

The problem defines a relation RR on the set (A×B,A×B)(A \times B, A \times B). An element of RR is an ordered pair of ordered pairs, denoted as ((a1,b1),(a2,b2))((a_1, b_1), (a_2, b_2)), where (a1,b1)A×B(a_1, b_1) \in A \times B and (a2,b2)A×B(a_2, b_2) \in A \times B. The condition for an element to be in RR is that a1a_1 divides b2b_2 and a2a_2 divides b1b_1.

Given sets: A={2,3,4}A = \{2, 3, 4\} B={8,9,12}B = \{8, 9, 12\}

The relation RR is given by: R={((a1,b1),(a2,b2))(A×B,A×B):a1b2 and a2b1}R = \{((a_1, b_1), (a_2, b_2)) \in (A \times B, A \times B) : a_1 | b_2 \text{ and } a_2 | b_1\}

Here, a1Aa_1 \in A, b1Bb_1 \in B, a2Aa_2 \in A, and b2Bb_2 \in B.

We need to count the number of quadruplets (a1,b1,a2,b2)(a_1, b_1, a_2, b_2) that satisfy the given conditions. The conditions a1b2a_1 | b_2 and a2b1a_2 | b_1 are independent of each other because the variables involved in one condition are distinct from the variables involved in the other condition. We can therefore use the Multiplication Principle.

Step 1: Count the number of pairs (a1,b2)(a_1, b_2) such that a1Aa_1 \in A, b2Bb_2 \in B, and a1b2a_1 | b_2.

We systematically check each element of AA against each element of BB for divisibility.

  • For a1=2a_1 = 2:
    • 282 | 8 (True)
    • 292 | 9 (False)
    • 2122 | 12 (True) This gives pairs (2,8)(2, 8) and (2,12)(2, 12).
  • For a1=3a_1 = 3:
    • 383 | 8 (False)
    • 393 | 9 (True)
    • 3123 | 12 (True) This gives pairs (3,9)(3, 9) and (3,12)(3, 12).
  • For a1=4a_1 = 4:
    • 484 | 8 (True)
    • 494 | 9 (False)
    • 4124 | 12 (True) This gives pairs (4,8)(4, 8) and (4,12)(4, 12).

The total number of pairs (a1,b2)(a_1, b_2) satisfying a1b2a_1 | b_2 is 2+2+2=62 + 2 + 2 = 6. Let N1N_1 be this count, so N1=6N_1 = 6.

Step 2: Count the number of pairs (a2,b1)(a_2, b_1) such that a2Aa_2 \in A, b1Bb_1 \in B, and a2b1a_2 | b_1.

This is structurally identical to Step 1, as the sets AA and BB are the same, and the divisibility condition is the same. We are choosing a2a_2 from AA and b1b_1 from BB.

  • For a2=2a_2 = 2: 282 | 8, 2122 | 12. Pairs: (2,8),(2,12)(2, 8), (2, 12).
  • For a2=3a_2 = 3: 393 | 9, 3123 | 12. Pairs: (3,9),(3,12)(3, 9), (3, 12).
  • For a2=4a_2 = 4: 484 | 8, 4124 | 12. Pairs: (4,8),(4,12)(4, 8), (4, 12).

The total number of pairs (a2,b1)(a_2, b_1) satisfying a2b1a_2 | b_1 is 2+2+2=62 + 2 + 2 = 6. Let N2N_2 be this count, so N2=6N_2 = 6.

Step 3: Calculate the total number of elements in RR using the Multiplication Principle.

An element of RR is determined by choosing a pair (a1,b2)(a_1, b_2) satisfying a1b2a_1 | b_2 and a pair (a2,b1)(a_2, b_1) satisfying a2b1a_2 | b_1. Since these choices are independent, the total number of elements in RR is the product of the number of ways for each independent choice.

Number of elements in R=(Number of ways to choose (a1,b2) such that a1b2)×(Number of ways to choose (a2,b1) such that a2b1)R = (\text{Number of ways to choose } (a_1, b_2) \text{ such that } a_1 | b_2) \times (\text{Number of ways to choose } (a_2, b_1) \text{ such that } a_2 | b_1) R=N1×N2|R| = N_1 \times N_2 R=6×6=36|R| = 6 \times 6 = 36.

3. Common Mistakes & Tips

  • Confusing the structure of elements: Remember that elements of RR are pairs of ordered pairs, i.e., ((a1,b1),(a2,b2))((a_1, b_1), (a_2, b_2)), not simple ordered pairs.
  • Independence of conditions: The conditions a1b2a_1 | b_2 and a2b1a_2 | b_1 involve disjoint sets of variables (a1,b2a_1, b_2 vs. a2,b1a_2, b_1), making them independent. This is key to applying the multiplication principle.
  • Systematic enumeration: For divisibility problems with small sets, listing all valid pairs is a robust method to ensure no possibilities are missed or double-counted.

4. Summary

The problem requires us to count the number of elements in a relation RR, which are quadruplets (a1,b1,a2,b2)(a_1, b_1, a_2, b_2) satisfying a1Aa_1 \in A, b1Bb_1 \in B, a2Aa_2 \in A, b2Bb_2 \in B, and the conditions a1b2a_1 | b_2 and a2b1a_2 | b_1. By observing that the two divisibility conditions are independent, we can count the number of ways each condition can be met separately and then multiply these counts. We found there are 6 pairs (a1,b2)(a_1, b_2) where a1Aa_1 \in A divides b2Bb_2 \in B, and 6 pairs (a2,b1)(a_2, b_1) where a2Aa_2 \in A divides b1Bb_1 \in B. Thus, the total number of elements in RR is 6×6=366 \times 6 = 36. This corresponds to option (C).

The final answer is 36\boxed{\text{36}}.

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