Given n(U)=140, n(A)=60, n(B)=55, n(C)=50, n(A∩B)=24, n(B∩C)=21, n(C∩A)=19, and n(A∩B∩C)=8, how many elements are in none of the sets?
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Answer and explanation
Correct answer: 31
Use the inclusion–exclusion principle for three sets: n(A∪B∪C) = n(A)+n(B)+n(C)−n(A∩B)−n(B∩C)−n(C∩A)+n(A∩B∩C). Substitution gives 60+55+50−24−21−19+8 = 109. The elements in none of the sets are outside the union, so their number is n(U)−n(A∪B∪C) = 140−109 = 31. Hence option A is correct.
Frequently asked questions
What is the correct answer to this question?
31
Why is this the correct answer?
Use the inclusion–exclusion principle for three sets: n(A∪B∪C) = n(A)+n(B)+n(C)−n(A∩B)−n(B∩C)−n(C∩A)+n(A∩B∩C). Substitution gives 60+55+50−24−21−19+8 = 109. The elements in none of the sets are outside the union, so their number is n(U)−n(A∪B∪C) = 140−109 = 31. Hence option A is correct.
Which subject and chapter does this question cover?
This is a Class 10 Mathematics question. Chapter: Sets. Topic: Venn Diagrams.
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