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. (i) A = {a, c, e, f, h}, B = {c, d, e, f} and C = {a, b, c, f} (ii) A = {1, 3, 5} B = {2, 3, 5, 6} and C = {1, 5, 6, 7}.
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Answer:
n(A ∪ B ∪ C) = n(A) + n(B) + n(C) – n(A ∩ B) – n(B ∩ C) – n(A ∩ C) + n(A ∩ B ∩ C) (i) A = {a, c, e, f, h}, B = {c, d, e, f}, C = {a, b, c, f} n (A) = 5, n (B) = 4, n (C) = 4 n( A ∩ B) = 3 n(B ∩ C) = 2 n( A ∩ C) = 3 n( A ∩ B ∩ C) = 2 A ∩ B = {c, e, f} B ∩ C = {c, f} A ∩ C = {a, c, f} A ∩ B ∩ C = {c, f} A ∪ B ∪ C = {a, c, d, e, f, b, h} ∴ 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) ∴ (1) = (2) ⇒ n(A ∪ B ∪ C) = n(A) + n(B) + n(C) – n(A ∩ B) – n(B ∩ C) – n(A ∩ C) + n(A ∩ B ∩ C) Hence it is verified. (ii) A = {1, 3, 5}, B = {2, 3, 5, 6 }, C = {1, 5, 6,7} = 3 n (B) = 4, n (C) = 4 n(A ∩ B) = 2 n(B ∩ C) = 2 n(C ∩ A) = 2 n(A ∩ B ∩ C) = 1 n(A ∪ B ∪ C) = 6 n(A ∪ B ∪ C) = n(A) + n(B) + n(C) – n(A ∩ B) – n(B ∩ C) – n(A ∩ C) + n(A ∩ B ∩ C) 6 = 3 + 4 + 4 – 2 – 2 – 2 + 1 = 12 – 6 = 6 Hence it is verified
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