यदि (n(A)=45), (n(B)=52), (n\(A\cup B\)=73) है, तो (n\(A\triangle B\)) कितना होगा, जहाँ \(A\triangle B=(A-B)\cup(B-A)\)?
If (n(A)=45), (n(B)=52), (n\(A\cup B\)=73), then what is (n\(A\triangle B\)), where \(A\triangle B=(A-B)\cup(B-A)\)?
Explanation opens after your attempt
A. (49)
Concept
First (n\(A\cap B\)=45+52-73=24), then (n\(A\triangle B\)=73-24=49). The symmetric difference excludes the common part.
Why this answer is correct
The correct answer is A. (49). First (n\(A\cap B\)=45+52-73=24), then (n\(A\triangle B\)=73-24=49). The symmetric difference excludes the common part.
Exam Tip
पहले (n\(A\cap B\)=45+52-73=24), फिर (n\(A\triangle B\)=73-24=49) है। सममित अंतर में साझा भाग शामिल नहीं होता।
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