Operations on Sets (Union, Intersection, Difference)

The set A is divided into its A-only part and its common part. Hence n(A ∩ B) = n(A) − n(A \ B) = 21 − 13 = 8. The union contains three disjoint parts: A \ B, A ∩ B, and B \ A. Therefore, n(A ∪ B) = 13 + 8 + 10 = 31. Option B is correct; 23 omits the common part, while 34 and 44 use unsuitable totals.

For two finite sets, the inclusion-exclusion principle states n(A ∪ B) = n(A) + n(B) − n(A ∩ B). Adding 14 and 10 counts each of the three common elements twice, so one copy of the intersection must be subtracted. Therefore, n(A ∪ B) = 14 + 10 − 3 = 21. Option B is correct; 24 fails to remove overlap, while 17 and 27 use incorrect arithmetic or operations.

For two finite sets, the inclusion–exclusion formula is \(n(A\cup B)=n(A)+n(B)-n(A\cap B)\). Therefore, \(n(A\cap B)=19+13-25=7\). The intersection is the part counted in both sets, so it must be subtracted once from the sum of the two set sizes. Hence, option B is correct. The values 5 and 12 result from incorrect arithmetic or using the formula in the wrong direction, while 32 is greater than the union and cannot be the intersection.

The set A consists of two disjoint parts: the elements only in A and the elements shared by A and B. Hence, \(n(A)=n(A\text{ only})+n(A\cap B)\). Therefore, the number only in A is \(23-9=14\). Option B is correct. The value 9 counts only the common region, 23 counts all elements of A including the common region, and 32 results from adding instead of subtracting the shared elements.

The total set B is divided into the region only in B and the common region A∩B. Therefore n(B)=n(B−A)+n(A∩B). Substituting the given values gives n(B−A)=27−11=16. Thus option B is correct. The value 11 is the overlap, 27 is all of B, and 38 is greater than the total number of elements in B.

By definition, A−B={x:x∈A and x∉B}. In a two-set Venn diagram, this is the portion of circle A that lies outside circle B, excluding their overlap. Therefore option A is correct. The overlap is A∩B, the B-only region belongs to B−A, and U includes regions unrelated to A−B.

The difference B−A contains elements that belong to B but do not belong to A. Consequently, on a Venn diagram it is the B-only region, obtained by removing the overlap A∩B from circle B. Option B is correct. The overlap is the intersection, A∪B includes both circles, and the A-only region represents A−B instead.

The intersection A ∩ B contains exactly the elements that occur in both sets. Comparing the members, 3 appears in A and B, and 5 also appears in A and B. The number 7 occurs only in B, while 1, 2, and 4 occur only in A. Hence A ∩ B = {3, 5}. Option D is the union, not the intersection.

The difference A − B consists of elements that are in A but not in B. The elements 2 and 8 are common to both sets, so they must be removed from A. The remaining elements, 4 and 6, belong only to A. Thus A − B = {4, 6}. Option B is the intersection, option C contains elements only in B, and option D is the complete set A.

To find B − A, retain the elements of B and remove every element that also belongs to A. Set B contains b, c, d, and e; c and e are common with A, so they are removed. The elements left only in B are b and d. Therefore B − A = {b, d}. Option A is the common part, while option D is the union of both sets.

The set A is divided into two non-overlapping parts: elements only in A, represented by A − B, and elements common to A and B, represented by A ∩ B. Therefore n(A) = n(A − B) + n(A ∩ B) = 10 + 7 = 17. Option B counts only the exclusive part and ignores the common elements, so it is incomplete.

Every element of B belongs either to the part only in B, B − A, or to the common part A ∩ B. These two regions do not overlap, so their cardinalities can be added: n(B) = n(B − A) + n(A ∩ B) = 13 + 6 = 19. Thus option C is correct. Option B omits the six common elements.

Let M be the set of students who like mathematics and S the set who like science. Students who like at least one subject belong to M ∪ S. By inclusion–exclusion, n(M ∪ S) = n(M) + n(S) − n(M ∩ S) = 24 + 18 − 7 = 35. The 7 students are subtracted once because they were counted twice.

The symbol ∅ means that A and B are disjoint, so their intersection has zero elements. The union formula is n(A ∪ B) = n(A) + n(B) − n(A ∩ B). Substituting the values gives 15 + 18 − 0 = 33. Since no element is counted in both sets, the two cardinalities can be added directly.

The relation A ⊆ B means that every element of A is already an element of B. When two sets have this relationship, taking their union adds no new elements to B; therefore, A ∪ B = B. Consequently, n(A ∪ B) = n(B) = 26. Option C is correct. The value 35 incorrectly adds the two cardinalities without removing the overlap.

Since A ⊆ B, every element of A is also an element of B. Thus, the elements common to A and B are exactly the elements of A, so A ∩ B = A. Therefore, n(A ∩ B) = n(A) = 12. Option A is correct. The intersection cannot have more elements than the smaller set, which also rules out 31 and 43.

Use the three-set inclusion–exclusion formula: n(A ∪ B ∪ C) = n(A) + n(B) + n(C) − n(A ∩ B) − n(A ∩ C) − n(B ∩ C) + n(A ∩ B ∩ C). Substitution gives 8 + 7 + 6 − 2 − 1 − 3 + 1 = 16. The triple intersection is added once because it was over-subtracted.

Set A is divided into two non-overlapping parts: the elements only in A, represented by A\B, and the elements common to A and B, represented by A∩B. Hence n(A)=n(A\B)+n(A∩B). Substituting the given values gives 20=13+n(A∩B), so n(A∩B)=20−13=7. Thus option A is correct; 13 is the exclusive part, not the common part.

The set B consists of its exclusive part, containing elements in B but not A, and its common part, A∩B. Therefore n(B)=n(B\A)+n(A∩B). The given values give 30=21+n(A∩B), so n(A∩B)=30−21=9. Option B is correct. The value 21 counts only-B elements, while 30 counts all elements of B, so neither is the intersection.

When A=B, both symbols refer to exactly the same set. Therefore A∩B=A∩A. By the idempotent law of intersection, X∩X=X, so A∩B=A. The empty set would result only for disjoint sets, and U or A′ cannot be selected without additional conditions. Thus option A is the only generally valid answer.