If n(A) = 55, n(B) = 60, n(C) = 65, n(A ∪ B ∪ C) = 120, and n(A ∩ B) + n(B ∩ C) + n(C ∩ A) = 78, what is n(A ∩ B ∩ C)?
Answer and explanation
Correct answer: 18
The governing concept is the inclusion–exclusion principle for three finite sets. It states n(A∪B∪C)=n(A)+n(B)+n(C)−[n(A∩B)+n(B∩C)+n(C∩A)]+n(A∩B∩C). Let x=n(A∩B∩C). Substitution gives 120=55+60+65−78+x. The sum of the individual cardinalities is 180, and 180−78=102, so 120=102+x. Therefore x=18. Option A is correct. The triple intersection is added back because elements common to all three sets are subtracted too many times during the pairwise subtraction. The other numerical choices do not satisfy the inclusion–exclusion equation.
Frequently asked questions
What is the correct answer to this question?
18
Why is this the correct answer?
The governing concept is the inclusion–exclusion principle for three finite sets. It states n(A∪B∪C)=n(A)+n(B)+n(C)−[n(A∩B)+n(B∩C)+n(C∩A)]+n(A∩B∩C). Let x=n(A∩B∩C). Substitution gives 120=55+60+65−78+x. The sum of the individual cardinalities is 180, and 180−78=102, so 120=102+x. Therefore x=18. Option A is correct. The triple intersection is added back because elements common to all three sets are subtracted too many times during the pairwise subtraction. The other numerical choices do not satisfy the inclusion–exclusion equation.
Which subject and chapter does this question cover?
This is a Class 10 Mathematics question. Chapter: Sets. Topic: Venn Diagrams.
