Notes On Application of Sets - CBSE Class 11 Maths
Results based on intersection, union and difference of two sets. 1) n(A ∪ B) = n(A) + n(B) A ∩ B = ∅ Example: Consider two sets A and B. A = {1,2,3,4} B = {5,6,7,8,9} A ∩ B = ∅ n(A) = 4 n(B) = 5 A ∪ B = {1,2,3,4,5,6,7,8,9} n(A ∪ B) = 9 n(A ∪ B) = n(A) + n(B) ⇒ 9 = 4 + 5 = 9 Therefore, the relation is verified. 2) n(A ∪ B) = n(A) + n(B) - n(A ∩ B) A and B are finite sets. Verification of the relation with the help of a Venn diagram: It can be observed that, (A - B) ∩ (A ∩ B) ∩ (B - A) = ∅ (A - B) ∩ (A ∩ B) ∪ (B - A) = A ∪ B n(A ∪ B) = n(A - B) + n(A ∩ B) + n(B - A) n(A ∪ B) = n(A - B) + n(A ∩ B) + n(B - A) + n(A ∩ B) - n(A ∩ B) n(A ∪ B) = n(A) + n(B) - n(A ∩ B) Hence, the relation is verified. 3) n(A ∪ B ∪ C) = n(A) + n(B) + n(C) - n(A ∩ B) - n(B ∩ C) - n(A ∩ C) + n(A ∩ B ∩ C) n(A ∪ D) = n(A) + n(D) - n(A ∩ D) n(A ∪ B ∪ C) = n(A) + n(B ∪ C) - n(A ∩ (B ∪ C)) n(A ∪ B ∪ C) = n(A) + n(B ∪ C) - n((A ∩ B) ∪ (A ∩ C)) n(A ∪ B ∪ C) = n(A) + n(B) + n(C) - n(B ∩ C) - n(A ∩ B) - n(A ∩ C) + n(A ∩ B ∩ C) .

#### Summary

Results based on intersection, union and difference of two sets. 1) n(A ∪ B) = n(A) + n(B) A ∩ B = ∅ Example: Consider two sets A and B. A = {1,2,3,4} B = {5,6,7,8,9} A ∩ B = ∅ n(A) = 4 n(B) = 5 A ∪ B = {1,2,3,4,5,6,7,8,9} n(A ∪ B) = 9 n(A ∪ B) = n(A) + n(B) ⇒ 9 = 4 + 5 = 9 Therefore, the relation is verified. 2) n(A ∪ B) = n(A) + n(B) - n(A ∩ B) A and B are finite sets. Verification of the relation with the help of a Venn diagram: It can be observed that, (A - B) ∩ (A ∩ B) ∩ (B - A) = ∅ (A - B) ∩ (A ∩ B) ∪ (B - A) = A ∪ B n(A ∪ B) = n(A - B) + n(A ∩ B) + n(B - A) n(A ∪ B) = n(A - B) + n(A ∩ B) + n(B - A) + n(A ∩ B) - n(A ∩ B) n(A ∪ B) = n(A) + n(B) - n(A ∩ B) Hence, the relation is verified. 3) n(A ∪ B ∪ C) = n(A) + n(B) + n(C) - n(A ∩ B) - n(B ∩ C) - n(A ∩ C) + n(A ∩ B ∩ C) n(A ∪ D) = n(A) + n(D) - n(A ∩ D) n(A ∪ B ∪ C) = n(A) + n(B ∪ C) - n(A ∩ (B ∪ C)) n(A ∪ B ∪ C) = n(A) + n(B ∪ C) - n((A ∩ B) ∪ (A ∩ C)) n(A ∪ B ∪ C) = n(A) + n(B) + n(C) - n(B ∩ C) - n(A ∩ B) - n(A ∩ C) + n(A ∩ B ∩ C) .

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