Let A={2,3,4} and B={8,9,12}. Then the number of elements in the relation R={((a₁, b₁),(a₂, b₂)) ∈(A × B, A × B): a₁ divides b₂ and a₂ divides b₁} is :

Solution

Maths sir · 3 steps

The two conditions are independent, so count the good pairs (a, b) once and square that count.

1. Split the conditions. An element of R is a choice of four numbers a₁, b₁, a₂, b₂. The rule pairs a₁ with b₂ and pairs a₂ with b₁, and these two conditions share no variable. So the count is (number of pairs (a₁, b₂) with a₁ | b₂) times (number of pairs (a₂, b₁) with a₂ | b₁).

   a ∈ A	Elements of B that a divides	Count
   2	8, 12	2
   3	9, 12	2
   4	8, 12	2

2. One condition. So there are 2 + 2 + 2 = 6 pairs (a, b) with a ∈ A, b ∈ B and a | b. Both (a₁, b₂) and (a₂, b₁) must be one of these 6 pairs.

   n(R) = 6 × 6 = 36

3. So. The relation has 36 elements, which is option (c).

The idea to keep. When a counting condition splits into parts that share no variable, count each part on its own and multiply. Here the cross-matching of a₁ with b₂ is only a relabelling.

Answer from the NTA final answer key. Solution by PrepWiser.