Find the number of elements in the union of 4 sets A, B, C and D having 150, 180, 210 and 240 elements respectively, given that each pair of sets has 15 elements in common. Each triple of sets has 3 elements in common and A ∩ B ∩ C ∩ D = ϕ 

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.