Let A={1,3,4,6,9} and B={2,4,5,8,10}. Let R be a relation defined on A × B such that R={((a₁, b₁),(a₂, b₂)): a₁ ≤ b₂ and b₁ ≤ a₂}. Then the number of elements in the set R is :

Solution

Maths sir · 2 steps

The two conditions never share a variable, so count each one separately and multiply.

1. Split the conditions. An element of R is a choice of four numbers a₁, a₂ ∈ A and b₁, b₂ ∈ B. The condition a₁ ≤ b₂ involves only a₁, b₂. The condition b₁ ≤ a₂ involves only b₁, a₂. The two choices are independent, so the counts multiply.

   a₁	b₂ ≥ a₁	b₁	a₂ ≥ b₁
   1	2, 4, 5, 8, 10: 5	2	3, 4, 6, 9: 4
   3	4, 5, 8, 10: 4	4	4, 6, 9: 3
   4	4, 5, 8, 10: 4	5	6, 9: 2
   6	8, 10: 2	8	9: 1
   9	10: 1	10	none: 0
   Total	16	Total	10

   n(R) = 16 × 10 = 160

2. So. There are 16 good (a₁, b₂) pairs and 10 good (b₁, a₂) pairs. Any one of the first can go with any one of the second, so R has 160 elements, option (d).

The idea to keep. When a counting condition breaks into parts that share no variable, count each part on its own and multiply. Add only when the cases cannot happen together.

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