यदि (n\(A\cup B\cup C\)=184), (n(A)=86), (n(B)=91), (n(C)=88) और (n\(A\cap B\cap C\)=23) है, तो (n\(A\cap B\)+n\(B\cap C\)+n\(C\cap A\)) कितना है?
If (n\(A\cup B\cup C\)=184), (n(A)=86), (n(B)=91), (n(C)=88) and (n\(A\cap B\cap C\)=23), then what is (n\(A\cap B\)+n\(B\cap C\)+n\(C\cap A\))?
Explanation opens after your attempt
A. (104)
Concept
Using the formula (184=86+91+88-S+23), so (S=104). Here (S) is the sum of the three pairwise intersections.
Why this answer is correct
The correct answer is A. (104). Using the formula (184=86+91+88-S+23), so (S=104). Here (S) is the sum of the three pairwise intersections.
Exam Tip
सूत्र से (184=86+91+88-S+23), इसलिए (S=104) है। यहाँ (S) तीनों युग्म प्रतिच्छेदों का योग है।
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