Venn Diagrams

The union A ∪ B contains elements in A only, B only, and both sets. To retain the elements in exactly one set, remove the common intersection from the union. Thus, n(exactly one)=n(A ∪ B)−n(A ∩ B)=63−17=46. Equivalently, the exactly-one region is (A−B)∪(B−A), so option A is correct.

Two sets are called disjoint when they have no element in common. The intersection records exactly the common elements, so for disjoint sets the intersection must be the empty set: \(A\cap B=\varnothing\). Their union need not be empty because both sets may contain many elements. Disjointness also does not imply equality or that one set is a subset of the other.

The expression \((A\cap B)^c\) is the complement of the intersection. First, \(A\cap B\) is the overlapping region common to A and B. Taking its complement means selecting every point in the universal set U outside that overlap. Thus the answer includes the A-only region, the B-only region, and the region outside both sets, but excludes only the common overlap.

De Morgan’s first law states that the complement of a union equals the intersection of the complements: \((A\cup B)^c=A^c\cap B^c\). An element is outside \(A\cup B\) precisely when it is outside both A and B. Therefore, the correct answer is option A. In a Venn diagram, this is the region outside both circles.

De Morgan’s second law states that the complement of an intersection equals the union of the complements: \((A\cap B)^c=A^c\cup B^c\). An element is outside the common part of A and B if it is outside A, outside B, or outside both. Hence option A is correct. The intersection and union interchange when a complement is taken.

For any finite set X contained in the universal set U, the number of elements in its complement is \(n(X^c)=n(U)-n(X)\). Here, take \(X=A\cap B\). Therefore, \(n((A\cap B)^c)=75-18=57\). The value 18 is the intersection itself, while 75 is the entire universal set. Hence option A is correct.

Students choosing only Hindi are \(36-20=16\), because the 20 students who chose both have to be removed. Students choosing only English are \(42-20=22\). Thus, the number choosing exactly one subject is \(16+22=38\). Equivalently, use \(n(A)+n(B)-2n(A\cap B)=36+42-40=38\). Therefore, option A is correct.

First find the number who like at least one sport using inclusion–exclusion: \(n(C\cup F)=n(C)+n(F)-n(C\cap F)=31+29-11=49\). The people who like neither sport are outside this union. Hence, neither \(=65-49=16\). Therefore, option A is correct. Subtracting the overlap once prevents double-counting those who like both sports.

The set B can be divided into two disjoint regions: the common part \(A\cap B\) and the B-only part \(B-A\). Also, the union consists of A together with the B-only part, so \(n(A\cup B)=n(A)+n(B-A)=46+26=72\). Since this already equals the given union, no additional common part is present; therefore \(n(A\cap B)=0\).

The universal set contains every disjoint region shown in the two-set Venn diagram: the A-only region has 18 elements, the intersection has 12, the B-only region has 22, and the outside region has 8. Adding all regions gives \(n(U)=18+12+22+8=60\). Therefore, option A is correct.

A two-set Venn diagram has three disjoint regions inside the union: the part only in \(A\), the intersection, and the part only in \(B\). These are \(A-B\), \(A\cap B\), and \(B-A\), respectively. Therefore, \(n(A\cup B)=14+9+17=40\). The elements outside both sets are in the complement of the union, so their number is \(n(U)-n(A\cup B)=55-40=15\). Hence option A is correct. The value 40 is the size of the union, not the outside region.

Each pairwise intersection includes the central three-set intersection, so subtract 4 from each pair to obtain the pair-only regions. For A and B, the count is \(12-4=8\); for B and C, it is \(13-4=9\); and for C and A, it is \(10-4=6\). Therefore, exactly two sets contain \(8+9+6=23\) elements. Option A is correct.

For three sets, use the inclusion–exclusion formula: n(A∪B∪C)=n(A)+n(B)+n(C)−n(A∩B)−n(B∩C)−n(C∩A)+n(A∩B∩C). Thus, n(A∪B∪C)=25+22+19−7−6−5+2=50. The elements in none of the sets are outside the union, so n(U)−n(A∪B∪C)=60−50=10. Therefore, option A is correct.

The wording “only” is important: each of the first six values represents a separate Venn-diagram region, and the final value represents the central region common to all three sets. Every element in the union belongs to exactly one of these seven disjoint regions. Therefore, n(A∪B∪C)=11+13+9+5+4+6+3=51. Hence option A is correct.

The expression A ∩ Bᶜ means the elements that belong to A and do not belong to B. The region common to A and the complement of B is therefore the part of A lying outside B. By the definition of set difference, this region is A − B. It is not B − A, because that would represent elements in B but outside A.

First find the union from its complement: n(A ∪ B) = n(U) − n((A ∪ B)ᶜ) = 95 − 20 = 75. Then apply inclusion–exclusion, n(A ∪ B) = n(A) + n(B) − n(A ∩ B). Thus 75 = 48 + 39 − n(A ∩ B), so n(A ∩ B) = 87 − 75 = 12. Option A is correct; the complement value 20 is not the intersection.

For two finite sets, the union formula is n(A∪B)=n(A)+n(B)−n(A∩B). The intersection is subtracted because elements common to A and B are counted once in n(A) and once again in n(B). Therefore, n(A∪B)=46+38−16=68. The universal-set size is not needed for this calculation. Thus, option B is correct.

Use the two-set inclusion–exclusion identity n(A∪B)=n(A)+n(B)−n(A∩B). Rearranging gives n(A∩B)=n(A)+n(B)−n(A∪B). Substitution yields 52+47−79=99−79=20. The common elements must be subtracted because adding the two set totals counts them twice. Therefore, option B is the unique correct answer.

First calculate the union using n(A∪B)=n(A)+n(B)−n(A∩B). Thus, n(A∪B)=61+54−25=90. The elements in neither A nor B are outside the union, so their number is n(U)−n(A∪B)=120−90=30. Equivalently, they form the complement of A∪B within U. Therefore, option C is correct.

The total n(A)=44 includes two parts: the region only in A and the common region A∩B. Therefore, the only-A region is found by subtracting the overlap from the total of A: n(A only)=n(A)−n(A∩B)=44−19=25. The value 19 is the intersection, while 44 includes both parts. Hence option B is correct.