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 = {a, c, e, f, h}, B = {c, d, e, f} and C = {a, b, c, f}

Answer 1

A ∩ B = {a, c, e, f, h} ∩ {c, d, e, f}

= {c, e, f}

B ∩ C = {c, d, e, f} ∩ {a, b, c, f}

= {c, f}

A ∩ C = {a, c, e, f, h} ∩ {a, b, c, f}

= {c, f}

(A ∩ B ∩ C) = {a, c, e, f, h} ∩ {c, d, e, f} ∩ {a, b, c, f}

= {c, f}

(A ∪ B ∪ C) = {a, c, e, f, h} ∪ {c, d, e, f} ∪ {a, b, c, f}

= {a, b, c, d, e, f, h}

n(A ∩ B) = 3, n(B ∩ C) = 2, n(A ∩ C) = 3, n(A ∩ B ∩ C) = 2

n(A ∪ B ∪ C) = 7  ...(1)

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

= 5 + 4 + 4 – 3 – 2 – 3 + 2

= 15 – 8

= 7 ...(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)