Let A={2,3,6,7} and B={4,5,6,8}. Let R be a relation defined on A × B by (a₁, b₁) R(a₂, b₂) if and only if a₁+a₂=b₁+b₂. Then the number of elements in R is __________.

Solution

Maths sir · 3 steps

Do not chase pairs of pairs. Group by the common sum s and multiply the ways A and B each make s.

1. What to count. An element of R is a choice of a₁, a₂ ∈ A and b₁, b₂ ∈ B with a₁+a₂=b₁+b₂. Order matters, so (a₁,a₂)=(2,3) and (3,2) are different choices.
2. Count by the sum. For each sum s, let n_A(s) be the number of ordered pairs from A adding to s, and n_B(s) the same for B. Every A-pair can go with every B-pair that has the same sum.

   n(R) = Σₛ n_A(s) n_B(s)

   Common sum s	Ordered pairs from A	Ordered pairs from B	Product
   8	(2,6),(6,2): 2	(4,4): 1	2
   9	(2,7),(7,2),(3,6),(6,3): 4	(4,5),(5,4): 2	8
   10	(3,7),(7,3): 2	(4,6),(6,4),(5,5): 3	6
   12	(6,6): 1	(4,8),(8,4),(6,6): 3	3
   13	(6,7),(7,6): 2	(5,8),(8,5): 2	4
   14	(7,7): 1	(6,8),(8,6): 2	2

3. So. The other sums (4, 5, 6 from A and 11, 16 from B) have no partner, so they add nothing. n(R) = 2+8+6+3+4+2 = 25.

The idea to keep. When a condition reads 'sum from one set equals sum from another', list the possible sums, count the ordered ways each side makes them, and add up the products.

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