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For 3 sets A , B & C

n(A) = Number of elements of set A

n(B) = Number of elements of set B

n(C) = Number of elements of set C

 

n(A ∪ B ∪ C) = n(A) + n(B) + n(C) – n(A ∩ B) – n(B ∩ C) – n(A ∩ C) + n(A ∩ B ∩ C)

 

Proof of n(A ∪ B ∪ C) Formula

We know that

P(E ∪ F) = P(E) + P(F) − P(E ∩ F).

 

Putting E = B, F = (B ∪ C)

P(A ∪ (B ∪ C)) = P(A) + P(B ∪ C) − P(A ∩ (B ∪ C)),

 

And also

P(A ∩ (B ∪ C)) = P((A ∩ B) ∪ (A ∩ C))

  = P(A ∩ B) + P(A ∩ C) − P(A ∩ B ∩ A ∩ C)

  = P(A ∩ B) + P(A ∩ C) − P(A ∩ B ∩ C)

and

P(B ∪ C) = P(B) + P(C) − P(B ∩ C).

 

Substituting the 2 nd and 3 rd equations into the first we get

P(A ∪ B ∪ C) = P(A) + P(B) + P(C) − P(B ∩ C) − P(A ∩ B) − P(A ∩ C) + P(A ∩ B ∩ C)

  1. Chapter 1 Class 11 Sets
  2. Concept wise

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.