Venn Diagrams

For each pairwise intersection, remove the elements that are also in the third set. Thus, elements only in A and B are 18−6=12, only in B and C are 16−6=10, and only in C and A are 14−6=8. These three disjoint regions contain exactly two-set members, so the total is 12+10+8=30. The triple intersection must be removed from every pair.

To count elements only in A, begin with all 50 elements of A. Subtract the 21 elements shared by A and B and the 19 shared by A and C. The 8 elements in all three sets were subtracted twice, so add them back once: 50−21−19+8=18. Therefore, 18 elements are in A but in neither B nor C.

The symmetric difference A △ B consists of the elements that are in A but not in B together with the elements that are in B but not in A. These two parts are disjoint, so n(A △ B) = n(A − B) + n(B − A). Substituting the given values gives 64 = 27 + n(B − A). Hence n(B − A) = 64 − 27 = 37, so option A is correct. The common elements of A and B are excluded from the symmetric difference.

Let x = n(A ∩ B). The elements that lie in exactly one of the two sets are the elements in A but not B plus the elements in B but not A. Their number is (39 − x) + (44 − x) = 83 − 2x. Since this number is 51, we obtain 83 − 2x = 51, so 2x = 32 and x = 16. Therefore, n(A ∩ B) = 16 and option B is correct. This is also consistent with the formula for symmetric difference.

Let M be the set of students who like Mathematics and P the set who like Physics. By the inclusion–exclusion principle, n(M union P) = n(M) + n(P) − n(M intersection P) = 34 + 31 − 13 = 52. Therefore, the number who like neither subject is the total number minus the union: 72 − 52 = 20. Hence option B is correct.

A union B is partitioned into exactly three mutually disjoint regions: the A-only region A − B, the common region A intersection B, and the B-only region B − A. Therefore, n(A union B) = n(A − B) + n(A intersection B) + n(B − A). Substituting the values gives 70 = 26 + n(A intersection B) + 19. Hence n(A intersection B) = 70 − 26 − 19 = 25. Option A is correct.

The complement of A union B contains the elements outside both A and B. Therefore, n(A union B) = n(U) − n((A union B) complement) = 100 − 18 = 82. For two finite sets, n(A union B) = n(A) + n(B) − n(A intersection B). Substituting gives 82 = 48 + 46 − n(A intersection B), so n(A intersection B) = 94 − 82 = 12. Thus option B is correct.

The union A ∪ B ∪ C consists of every region lying inside at least one of the three sets. Since the question gives the seven mutually exclusive interior regions, we add them once each: 14 + 18 + 16 + 7 + 6 + 5 + 4 = 70. The triple-intersection value is already one separate region, so it must not be added again through another formula.

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.

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). Let x=n(A∩B∩C). Substitution gives 82=40+36+34−12−10−9+x, or 82=79+x. Hence x=3. The common central region of all three sets therefore contains 3 elements, so option A is correct.

The difference A − B contains the elements that belong to A but not to B. The intersection A ∩ B contains the elements that belong to both A and B. These two regions are disjoint parts that together cover every element of A: an element of A is either outside B or inside B. Therefore, (A − B) ∪ (A ∩ B) = A.

The union A ∪ B contains every element in A or B. When all elements belonging to A are removed from this union, the remaining elements must be those that belong to B but do not belong to A. Hence, (A ∪ B) − A = B − A. In a Venn diagram, this is the non-overlapping portion of circle B.

The union of two sets is partitioned into three disjoint Venn regions: A−B, B−A, and A∩B. Therefore, n(A∪B) = n(A−B) + n(B−A) + n(A∩B). Substituting the data gives 58 = 22 + 20 + n(A∩B). Thus n(A∩B) = 58−42 = 16. The other choices result from using only one region or subtracting incorrectly, so option A is correct.

First calculate the union: n(A ∪ B) = n(A) + n(B) − n(A ∩ B) = 42 + 36 − 14 = 64. The elements in neither A nor B form the complement of A ∪ B in the universal set U. Hence their number is n(U) − n(A ∪ B) = 85 − 64 = 21. Therefore, option A is correct.

The total n(A) includes three kinds of elements: those only in A, those in A ∩ B but not C, and those in A ∩ C but not B; the triple intersection is included in both pairwise intersections. Thus the correct inclusion–exclusion expression is only A = n(A) − n(A ∩ B) − n(A ∩ C) + n(A ∩ B ∩ C). Hence only A = 38 − 16 − 13 + 5 = 14.

For two finite sets, the cardinality of the union is n(A ∪ B) = n(A) + n(B) − n(A ∩ B). The common 21 elements are present in both A and B, so they would be counted twice in 57 + 49 and must be subtracted once. Therefore, n(A ∪ B) = 57 + 49 − 21 = 85. The value of n(U) is not needed for this calculation.

First find the number in at least one set: n(A ∪ B) = n(A) + n(B) − n(A ∩ B) = 72 + 64 − 28 = 108. The people in neither set are outside this union but still inside U. Therefore, neither = n(U) − n(A ∪ B) = 140 − 108 = 32. Subtracting the intersection prevents the 28 common people from being counted twice.

The total n(A) consists of the elements only in A together with the elements in the overlap A ∩ B. Therefore, n(A only) = n(A) − n(A ∩ B) = 59 − 24 = 35. The value 24 represents the common region, while 59 represents all elements of A, including that common region. Thus option B is correct; 83 does not follow from the given cardinalities.

The universal set contains every region shown in the Venn diagram. These regions are the A-only region A − B, the common region A ∩ B, the B-only region B − A, and the region outside both sets. Therefore, n(U) = 26 + 18 + 33 + 12 = 89. Since the regions are disjoint and together cover U, no subtraction or overlap correction is needed.

The union A∪B consists of three disjoint regions: elements only in A, common elements, and elements only in B. Therefore, n(A∪B)=n(A−B)+n(A∩B)+n(B−A)=29+17+22=68. The universal set has 96 elements, so the outside region contains 96−68=28 elements. Hence option A is correct.

Since A is a subset of B, every element of A is already included in B. The difference B−A therefore contains the elements of B that are not in A. For finite sets, n(B−A)=n(B)−n(A)=82−34=48. Thus 48 elements belong exclusively to B, and option A is the only correct answer.

The union contains elements in A only, B only, and both sets. The elements belonging to exactly one set are obtained by removing the common region from the union: n(exactly one)=n(A∪B)−n(A∩B)=98−42=56. Thus option A is correct. The value 98 includes the intersection, while 42 counts only the common region.

For three sets, the inclusion–exclusion formula is n(A∪B∪C)=n(A)+n(B)+n(C)−n(A∩B)−n(B∩C)−n(C∩A)+n(A∩B∩C). Substituting the values gives 42+39+35−15−13−11+5=87. The triple intersection is added once because it was subtracted twice while removing the pairwise overlaps.

Each pairwise intersection includes the seven elements common to all three sets. Therefore, the regions belonging to exactly two sets are (21−7), (19−7), and (16−7). Their total is 14+12+9=35. The triple intersection must be subtracted separately from each pairwise intersection because it is included in all three given pair counts.

To count elements only in A, subtract the elements shared with B and the elements shared with C, then add the triple intersection once because it was subtracted twice: only A = n(A)−n(A∩B)−n(A∩C)+n(A∩B∩C). Thus, only A = 63−26−24+9=22. Therefore, option B is correct.