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