Power Set and Subsets

Correct answer: B. 128

Explanation: The natural numbers from 1 through 12 are the elements of U. The primes in this interval are 2, 3, 5, 7, and 11, so A has 5 elements. Therefore, its complement A' has 12 − 5 = 7 elements. The power set of a 7-element set has 2^7 = 128 members. Note that 1 is not prime. Hence option B is correct.

Correct answer: B. 16

Explanation: Because 1 must be included, one element is already fixed in every permitted subset. For the total cardinality to be even, the chosen number of elements from the remaining five elements {2,3,4,5,6} must be odd. Among five elements, the number of subsets with odd cardinality equals the number with even cardinality, namely 2⁵ ÷ 2 = 16. Therefore, option B is correct.

Correct answer: B. 4

Explanation: Use the inclusion–exclusion formula to find the union: |A ∪ B| = |A| + |B| − |A ∩ B| = 6 + 5 − 3 = 8. Since U has 10 elements, the complement of A ∪ B has 10 − 8 = 2 elements. A two-element set has 2² = 4 subsets in its power set. Thus, option B is correct.

Correct answer: B. 8

Explanation: The set A has 3 elements, so its power set P(A) has 2³ = 8 elements. In general, for any set S, the singleton members of P(S) are exactly {s} for each s ∈ S. Therefore, P(P(A)) has one singleton member for each member of P(A), giving 8 singleton sets: {∅}, {{x}}, {{y}}, {{z}}, {{x,y}}, {{x,z}}, {{y,z}}, and {{x,y,z}}. Hence, option B is correct.

Correct answer: C. 6

Explanation: If A has n elements, then its power set has 2^n elements. Since U has 12 elements, the complement A' has 12 - n elements, and therefore P(A') has 2^(12-n) elements. The given equality becomes 2^n = 2^(12-n). Equal powers of the same base have equal exponents, so n = 12 - n. Thus 2n = 12 and n = 6. Therefore, option C is correct.

Correct answer: B. 15

Explanation: For a finite set X, the power set has 2^|X| members. Since |𝒫(A)| = 128 = 2^7, A has 7 elements. The complement A′ is taken in U, so |A′| = |U| − |A| = 11 − 7 = 4. A four-element set has 2^4 = 16 subsets, including the empty set. Therefore, its non-empty subsets number 16 − 1 = 15, so option B is correct.

Correct answer: C. 64

Explanation: First find the intersection: A ∩ B = {3,4}, which has 2 elements. The complement is taken relative to U, so (A ∩ B)′ = U − {3,4} = {1,2,5,6,7,8}. This complement has 6 elements. The number of subsets of any six-element set is 2^6 = 64, including the empty subset. Hence |𝒫((A ∩ B)′)| = 64, making option C correct.

Correct answer: B. 8

Explanation: The fixed element a is already included, contributing one element. To make the complete subset have odd cardinality, we must select an even number of elements from the remaining four elements. Among four elements, the number of subsets with even size is 2^(4−1) = 8. Equivalently, half of all 2^4 choices have even size. Therefore, eight members of 𝒫(A) satisfy both conditions, so option B is correct.

Correct answer: C. 2^8

Explanation: For any finite set X, |𝒫(X)| = 2^|X|. Since |𝒫(A)| = 64 = 2^6, we obtain |A| = 6. Because A is a subset of U and |U| = 14, its complement has |A′| = 14 − 6 = 8 elements. Therefore the power set of A′ contains 2^8 subsets. Thus option C, 2^8, is correct.

Correct answer: C. A = B

Explanation: 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.

Correct answer: B. 64

Explanation: 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.

Correct answer: B. 16

Explanation: 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.

Correct answer: B. {{2,3}}

Explanation: 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.

Correct answer: B. 4

Explanation: 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.

Correct answer: B. 3

Explanation: 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.

Correct answer: C. 56

Explanation: 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.

Correct answer: C. 16

Explanation: 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.

Correct answer: C. 64

Explanation: 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.

Correct answer: A. Only {1} ∈ P(A)

Explanation: Because {1} is a subset of A = {1, 2}, it is an element of P(A); therefore {1} ∈ P(A) is true. However, {1} ⊆ P(A) means that the element 1 itself must belong to P(A). The elements of P(A) are ∅, {1}, {2}, and {1, 2}; the number 1 is not one of them. Thus only the first statement is true, so option A is correct.

Correct answer: B. 4095

Explanation: The primes from 1 through 20 are 2, 3, 5, 7, 11, 13, 17, and 19, so A has 8 elements. Since U has 20 elements, the complement A' has 20 − 8 = 12 elements. Its power set contains 2^12 = 4096 subsets, including the empty set. Therefore the number of non-empty subsets is 4096 − 1 = 4095, so option B is correct.

Correct answer: C. 20

Explanation: The odd elements form A = {1, 3, 5, 7, 9, 11}, so the complement A' consists of the six even elements {2, 4, 6, 8, 10, 12}. Members of P(A') are subsets of A'. The number of members containing exactly 3 elements is the number of ways to choose 3 elements from 6, namely C(6,3) = 6!/(3!3!) = 20. Thus option C is correct.

Correct answer: A. Always

Explanation: Since A is a subset of U, its complement has cardinality |A'| = |U| − |A| = 2m − m = m. The power set of any set with m elements has 2^m elements. Therefore |P(A)| = 2^m and |P(A')| = 2^m, so the two cardinalities are equal for every allowed value of m, not just for a special value. Hence option A is correct.

Correct answer: B. 255

Explanation: For a finite set A, the power set has 2^|A| members. Since |P(A)| = 256 = 2^8, we get |A| = 8. The universal set U has 16 elements, so its complement A' has |A'| = 16 − 8 = 8 elements. A set with 8 elements has 2^8 = 256 subsets, including the empty set. Therefore, the number of non-empty subsets is 256 − 1 = 255, so option B is correct.

Correct answer: A. 10

Explanation: We are counting 4-element subsets of A, not subsets of the power set as objects of a second-level selection. The element a must be included and b must be excluded. Thus, one of the four positions is already occupied by a, while the remaining three elements must be chosen from c,d,e,f,g, which gives five available elements. The number of choices is C(5,3) = 5!/(3!2!) = 10. Hence option A is correct.

Correct answer: B. 3

Explanation: Let |A| = n. The first power set P(A) has 2^n elements. Taking the power set again gives |P(P(A))| = 2^(2^n). The given value is 256 = 2^8, so 2^n = 8. Since 8 = 2^3, it follows that n = 3. Therefore |A| = 3, and option B is correct. The nested power set must be handled by applying the power-set formula twice.