यदि (n\(A\cup B\cup C\)=88), (n(A-B)=19), (n(B-A)=24), (n\(A\cap B\)=13) और (C-\(A\cup B\)) में (32) तत्व हैं, तो (n(\(A\cup B\)-C)) क्या है यदि (C) में \(A\cup B\) का कोई तत्व नहीं है?
If (n\(A\cup B\cup C\)=88), (n(A-B)=19), (n(B-A)=24), (n\(A\cap B\)=13), and (C-\(A\cup B\)) has (32) elements, what is (n(\(A\cup B\)-C)) if (C) has no element of \(A\cup B\)?
Explanation opens after your attempt
A. (,56,)
Concept
Since (C) and \(A\cup B\) are disjoint, (\(A\cup B\)-C=A\cup B). Its value is (19+24+13=56).
Why this answer is correct
The correct answer is A. (,56,). Since (C) and \(A\cup B\) are disjoint, (\(A\cup B\)-C=A\cup B). Its value is (19+24+13=56).
Exam Tip
क्योंकि (C) और \(A\cup B\) असंबद्ध हैं, इसलिए (\(A\cup B\)-C=A\cup B)। इसका मान (19+24+13=56) है।
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