Question

Verify n (A ∪ B ∪ C) = n(A) + n(B) + n(C) – n(A ∩ B) – (B ∩ C) – n(A ∩ C) + (A ∩ B ∩ C) for the following sets A = {1, 3, 5, 6, 8}, B = {3, 4, 5, 6} and C = {1, 2, 3, 6}

Answer 1

(A ∪ B ∪ C) = {1, 2, 3, 4, 5, 6, 8} 

n (A ∪ B ∪ C) = 7 

Also, n (A) = 5, n (B) = 4, n (C) = 4, 

Further, A ∩ B = {3, 5, 6} ⇒ n(A ∩ B) = 3 

B ∩ C = {3, 6} ⇒ n(B ∩ C) = 2 

A ∩ C = {3, 5, 6} ⇒ n(A ∩ C) = 3 

Also, A ∩ B ∩ C = {3, 6} ⇒ n(A ∩ B ∩ C) = 2

Now n (A ∪ B ∪ C) = n (A) + n (B) + n (C) – n(A ∩ B) – n( B ∩ C) – n (A ∩ C) + n(A ∩ B ∩ C) 

7 = 5 + 4 + 4 – 3 – 2 – 3 + 2 

7 = 13 – 8 + 2 

7 = 5 + 2 

7 = 7 

Thus verified