If \(n(A\cup B\cup C)=86\), \(n(A)=38\), \(n(B)=34\), \(n(C)=31\), \(n(A\cap B)=12\), \(n(B\cap C)=10\), and \(n(C\cap A)=9\), what is the value of \(n(A\cap B\cap C)\)?
Answer and explanation
Correct answer: 14
Use the inclusion–exclusion formula for three finite sets: \(n(A\cup B\cup C)=n(A)+n(B)+n(C)-n(A\cap B)-n(B\cap C)-n(C\cap A)+n(A\cap B\cap C)\). Substitution gives \(86=38+34+31-12-10-9+x=72+x\). Hence \(x=14\), so \(n(A\cap B\cap C)=14\). The triple intersection is added at the end because elements common to all three sets were counted three times initially and then subtracted three times through the pairwise terms.
Frequently asked questions
What is the correct answer to this question?
14
Why is this the correct answer?
Use the inclusion–exclusion formula for three finite sets: \(n(A\cup B\cup C)=n(A)+n(B)+n(C)-n(A\cap B)-n(B\cap C)-n(C\cap A)+n(A\cap B\cap C)\). Substitution gives \(86=38+34+31-12-10-9+x=72+x\). Hence \(x=14\), so \(n(A\cap B\cap C)=14\). The triple intersection is added at the end because elements common to all three sets were counted three times initially and then subtracted three times through the pairwise terms.
Which subject and chapter does this question cover?
This is a Class 10 Mathematics question. Chapter: Sets. Topic: Operations on Sets (Union, Intersection, Difference).
