यदि (n\(A\cup B\cup C\)=160), (n(A)=70), (n(B)=75), (n(C)=80) और (n\(A\cap B\cap C\)=18) है, तो (n\(A\cap B\)+n\(B\cap C\)+n\(C\cap A\)) कितना है?
If (n\(A\cup B\cup C\)=160), (n(A)=70), (n(B)=75), (n(C)=80) and (n\(A\cap B\cap C\)=18), then what is (n\(A\cap B\)+n\(B\cap C\)+n\(C\cap A\))?
Explanation opens after your attempt
A. (83)
Concept
Using the formula (160=70+75+80-S+18), so (S=83). Here (S) is the sum of the three pairwise intersections.
Why this answer is correct
The correct answer is A. (83). Using the formula (160=70+75+80-S+18), so (S=83). Here (S) is the sum of the three pairwise intersections.
Exam Tip
सूत्र से (160=70+75+80-S+18), इसलिए (S=83) है। यहाँ (S) तीनों युग्म प्रतिच्छेदों का योग है।
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