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Question
If A = {1, 2, 3, 4}, B = {3, 4, 5, 6}, C = {4, 5, 6, 7, 8} and universal set X = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}, then verify the following:
n(B) = (A' ∩ B) + n(A ∩ B)
Solution
A = {1, 2, 3, 4}, B = {3, 4, 5, 6}, C = {4, 5, 6, 7, 8},
X = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
The number of elements in B = {3, 4, 5, 6}
∴ n(B) = 4 .....(i)
The complement of A includes all elements in X that are not in A:
A' = X - A
A' = {5, 6, 7, 8, 9, 10}
The intersection of A′ and B includes elements that are in both A′ and B:
A' ∩ B = {5, 6}
∴ n (A' ∩ B) = 2
The intersection of A and B includes elements that are in both A and B:
A ∩ B = {3, 4}
∴ n (A ∩ B) = 2
∴ n (A' ∩ B) + n (A ∩ B) = 2 + 2 = 4 ...(ii)
From (i) and (ii), we get
n(B) = (A' ∩ B) + n(A ∩ B)
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