यदि (n(A)=55), (n(B)=60), (n(C)=65), (n\(A\cup B\cup C\)=120), और (n\(A\cap B\)+n\(B\cap C\)+n\(C\cap A\)=78) है, तो (n\(A\cap B\cap C\)) क्या है?
If (n(A)=55), (n(B)=60), (n(C)=65), (n\(A\cup B\cup C\)=120), and (n\(A\cap B\)+n\(B\cap C\)+n\(C\cap A\)=78), what is (n\(A\cap B\cap C\))?
Explanation opens after your attempt
A. (18)
Concept
Using the formula, (120=55+60+65-78+x), so (x=18). The sum of pairwise intersections is subtracted directly in the formula.
Why this answer is correct
The correct answer is A. (18). Using the formula, (120=55+60+65-78+x), so (x=18). The sum of pairwise intersections is subtracted directly in the formula.
Exam Tip
सूत्र से (120=55+60+65-78+x), इसलिए (x=18)। युग्म छेदन का योग सीधे सूत्र में घटता है।
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