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. (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}.

Answer 1

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

Step-by-step explanation:

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