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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}
Solution
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)
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