Correct Answer - Option 4 : 702
Calculation:
Given: The four sets have 150, 180, 210 and 240 elements respectively
n(A) = 150
n(B) = 180
n(C) = 210
n(D) = 240
Each pair of sets has 15 elements
n(A ∩ B) = 15
n(A ∩ C) = 15
n(A ∩ D) = 15
n(B ∩ C) = 15
n(B ∩ D) = 15
n(C ∩ D) = 15
Each triple of sets has 3 elements
n(A ∩ B ∩ C) = 3
n(A ∩ B ∩ D) = 3
n(A ∩ C ∩ D) = 3
n(B ∩ C ∩ D) = 3
A ∩ B ∩ C ∩ D = ϕ
n(A ∩ B ∩ C ∩ D) = 0
Now, number of elements in the union of 4 sets A, B, C and D
n(A ∪ B ∪ C ∪ D) = n(A) + n(B) + n(C) + n(D) - n(A ∩ B) - n(A ∩ C) - n(A ∩ D) - n(B ∩ C) - n(B ∩ D) - n(C ∩ D) + n(A ∩ B ∩ C) + n(A ∩ B ∩ D) + n(A ∩ C ∩ D) + n(B ∩ C ∩ D) - n(A ∩ B ∩ C ∩ D)
⇒ 150 + 180 + 210 + 240 - 6 × 15 + 4 × 3 - 0
∴ The required number of elements is 702.