Question

Verify n(A ∪ B ∪ C) = n(A) + n(B) + n(C) – n(A ∩ B) – n(B ∩ C) – n(A ∩ C) + n(A ∩ B ∩ C) for the following sets

A = {1, 3, 5}, B = {2, 3, 5, 6}, C = {1, 5, 6, 7}

Answer 1

A = {1, 3, 5}, B = {2, 3, 5, 6}, C = {1, 5, 6, 7}

A ∩ B = {1, 3, 5} ∩ {2, 3, 5, 6}

= {3, 5}

B ∩ C = {2, 3, 5, 6} ∩ {1, 5, 6, 7}

= {5, 6}

A ∩ C = {1, 3, 5} ∩ {1, 5, 6, 7}

= {1, 5}

A ∩ B ∩ C = {1, 3, 5} ∩ {2, 3, 5, 6} ∩ {1, 5, 6, 7}

= {5}

A ∪ B ∪ C = {1, 3, 5} ∪ {2, 3, 5, 6} ∪ {1, 5, 6, 7}

= {1, 2, 3, 5, 6, 7}

n(A) = 3, n(B) = 4, n(C) = 4

n(A ∩ B) = 2, n(B ∩ C) = 2, n(A ∩ C) = 2

n(A ∩ B ∩ C) = 1

n(A ∪ B ∪ C) = 6  ...(1)

n(A) + n(B) + n(C) – n(A ∩ B) – n(B ∩ C) – n(A ∩ C) + n(A ∩ B ∩ C)

= 3 + 4 + 4 – 2 – 2 – 2 + 1

= 12 – 6

= 6 ...(2)

From (1) and (2) we get

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