If \(n(A)=12\), \(n(B)=14\), \(n(C)=10\), \(n(A\cap B)=5\), \(n(A\cap C)=3\), \(n(B\cap C)=4\), and \(n(A\cap B\cap C)=2\), what is \(n(A\cup B\cup C)\)?
Answer and explanation
Correct answer: 26
For three finite sets, the inclusion–exclusion principle is \(n(A\cup B\cup C)=n(A)+n(B)+n(C)-n(A\cap B)-n(A\cap C)-n(B\cap C)+n(A\cap B\cap C)\). Substituting the given values gives \(12+14+10-5-3-4+2=26\). The pairwise intersections are subtracted because their elements were counted twice, and the triple intersection is added once because it was then subtracted too many times. Therefore, option B, 26, is correct.
Frequently asked questions
What is the correct answer to this question?
26
Why is this the correct answer?
For three finite sets, the inclusion–exclusion principle is \(n(A\cup B\cup C)=n(A)+n(B)+n(C)-n(A\cap B)-n(A\cap C)-n(B\cap C)+n(A\cap B\cap C)\). Substituting the given values gives \(12+14+10-5-3-4+2=26\). The pairwise intersections are subtracted because their elements were counted twice, and the triple intersection is added once because it was then subtracted too many times. Therefore, option B, 26, is correct.
Which subject and chapter does this question cover?
This is a Class 10 Mathematics question. Chapter: Sets. Topic: Operations on Sets (Union, Intersection, Difference).
