Power Set and Subsets

The union combines every element appearing in either set: A ∪ B = {1,2,3,4,5}. Taking the complement relative to U leaves (A ∪ B)′ = {6,7,8}, which has 3 elements. The power set of a three-element set has 2^3 = 8 members, including the empty subset. Therefore |𝒫((A ∪ B)′)| = 8, so option C is correct.

If 𝒫(A) = 𝒫(B), every subset of A is also a subset of B and every subset of B is also a subset of A. In particular, A is an element of 𝒫(A), so equality of the power sets gives A ∈ 𝒫(B), which means A ⊆ B. Similarly, B ∈ 𝒫(B) gives B ⊆ A. Mutual inclusion implies A = B. Thus option C is definitely true.

Because A′ contains the elements of U that are not in A, |A| = |U| − |A′| = 9 − 4 = 5. A set with n elements has 2ⁿ total subsets, so A has 2⁵ = 32 subsets. Proper subsets exclude the set itself, while still including the empty set; therefore, the number of proper subsets is 32 − 1 = 31. Hence, option B is correct.

The element 1 is compulsory, so there is no choice about including it. The element 2 is forbidden, so it also gives no choice and must be left out. The remaining elements are 3, 4, and 5, and each of these may either be included or excluded independently. Thus the number of valid subsets is 2³ = 8. Therefore, option B is correct.

The condition ‘neither p nor q’ means that both p and q must be excluded from every valid subset. The only elements available for selection are r, s, t, and u, giving four independently optional elements. Each can be either selected or not selected, so the number of subsets is 2⁴ = 16. This includes the empty subset. Hence, option B is correct.

For every nonempty finite set with n elements, exactly half of its 2^n subsets have even cardinality and half have odd cardinality. This follows by pairing each subset with the set obtained by toggling one fixed element. Here n=7, so the number of even-cardinality subsets is 2^7/2=2^6=64. Option B is correct.

Since 6 must be included, the selected subset already has one element. To make its total cardinality odd, the selection from the remaining five elements must contain an even number of elements. Among the subsets of a five-element set, exactly 2^4=16 have even cardinality. Therefore, option B is correct.

The elements of A are 1, the set {2,3}, and 4. A member of the power set must itself be a subset of A, meaning every one of its elements must be one of these three objects. The set {{2,3}} contains the single element {2,3}, which belongs to A, so it is a subset of A. Thus option B is correct.

Start with A = ∅, which has 0 elements. Its power set is P(A) = {∅}, so |P(A)| = 1. Taking the power set again gives a set with 2¹ = 2 elements, so |P(P(A))| = 2. Taking the power set a third time gives 2² = 4 elements. Therefore, |P(P(P(A)))| = 4, making option B correct.

Let |A| = n. A set with n elements has a power set of size 2ⁿ. Therefore, |P(P(A))| = 2^(2ⁿ). Since 256 = 2⁸, we have 2^(2ⁿ) = 2⁸, which gives 2ⁿ = 8. Because 8 = 2³, n = 3. It is important not to confuse 2^(2ⁿ) with 2^(2n); the exponent is itself 2ⁿ. Thus option B is correct.

Because A is a subset of B, every subset of A is also a subset of B; therefore P(A) is contained in P(B). The difference P(B)-P(A) contains all subsets of B except the 2^3 subsets of A. Since |P(B)|=2^6=64 and |P(A)|=2^3=8, the difference has 64-8=56 elements. Option C is correct.

A set belongs to both P(A) and P(B) exactly when it is a subset of both A and B. Being a subset of both means that it contains only elements common to A and B. Hence, P(A) ∩ P(B) = P(A ∩ B). Since |A ∩ B| = 4, its power set has 2⁴ = 16 members. Therefore, option C is correct.

Since A and B are disjoint, their only common subset is the empty set ∅. The power set of a set with n elements has 2ⁿ members, so |P(A)| = 2⁴ = 16 and |P(B)| = 2³ = 8. When taking the union, ∅ is counted twice and must be subtracted once. Therefore, |P(A) ∪ P(B)| = 16 + 8 − 1 = 23.

First calculate the difference: A − B = {1,7,9}, because 3 and 5 belong to both A and B. The prime symbol means complement relative to U. Hence (A − B)′ = U − {1,7,9} = {2,3,4,5,6,8}, which has 6 elements. Its power set therefore has 2⁶ = 64 members. Thus option C is correct.

“At least 6 elements” includes subsets having exactly 6 elements and exactly 7 elements. From a 7-element set, the number of 6-element subsets is C(7,6) = 7, because one element is omitted. The number of 7-element subsets is C(7,7) = 1, namely A itself. Therefore the total is 7 + 1 = 8, so option B is correct.

The set A has 8 elements. A subset with exactly r elements can be selected in C(8, r) ways. “At most 2 elements” includes subsets with 0, 1, or 2 elements. Therefore, the required number is C(8,0) + C(8,1) + C(8,2) = 1 + 8 + 28 = 37. The 0-element subset is the empty set, so it must be included.

There are two possibilities for the condition involving a and b: either a is included and b is excluded, or b is included and a is excluded. The remaining elements c, d, and e are unrestricted, so each can independently be included or excluded in 2³ ways. Hence the total number of subsets is 2 × 2³ = 16. Option B is correct.

The required subset must contain 1, so that element is already fixed. It must not contain 6, so 6 is excluded. To make a 3-element subset, two more elements must be selected from the remaining allowed set {2,3,4,5}. The number of choices is C(4,2) = 6. Therefore, option B is the correct answer.

A subset is disjoint from {1,2} if it contains neither 1 nor 2. Therefore, all such subsets must be formed from the remaining elements {3,4,5}. Every subset of {3,4,5} is allowed, including the empty set. Since this set has 3 elements, its power set has 2³ = 8 members. Hence option B is correct.

Every member of P(U) that contains A must include a, b, and c. The remaining elements d, e, f, g, and h are optional; each may be included or omitted independently. Thus the required subsets have the form A ∪ S, where S is any subset of U − A. Since U − A has 5 elements, the number of possibilities is 2⁵ = 32. Option C is correct.