Power Set and Subsets

Because A is a subset of B, the elements in B − A are obtained by subtracting the cardinality of A from that of B: |B − A| = 6 − 4 = 2. Also, since B is a subset of U, |U − B| = 9 − 6 = 3. A two-element set has 2^2 = 4 subsets, while a three-element set has 2^3 = 8 subsets. Hence the required sum is 4 + 8 = 12, so option B is correct.

The power set P(A) contains every subset of A. A singleton set is a set containing exactly one element. The singleton subsets of A are {1}, {2}, and {3}; there are no others because A has only three distinct elements. The empty set has zero elements and is not a singleton, while larger subsets contain more than one element. Therefore, the answer is 3, option C.

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.

A subset of A is disjoint from {1, 2} only when it contains neither 1 nor 2. Therefore, every such subset must be formed from the remaining elements {3, 4}. Each of these two elements may either be selected or not selected, giving 2^2 = 4 possible subsets: empty set, {3}, {4}, and {3, 4}. Hence option B is correct.

The complement of A with respect to U is A' = {4, 5, 6, 7}, so A' has 4 elements. A set with n elements has 2^n subsets, including the set itself. Therefore, A' has 2^4 = 16 total subsets. Proper subsets exclude the set A' itself, so their number is 16 - 1 = 15. Hence option C is correct.

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.

For any finite set X, the number of members in its power set is 2^|X|. Since |P(A)| = 32 = 2^5, A has 5 elements. The complement A' therefore has 15 - 5 = 10 elements because U has 15 elements. Hence |P(A')| = 2^10. Option A is the size of P(A), not P(A'), so option B is the only correct answer.

The set A has 5 elements, so its power set P(A) contains 2^5 = 32 subsets. We need subsets whose cardinality is greater than 2, namely those having 3, 4, or 5 elements. Their numbers are C(5,3) = 10, C(5,4) = 5, and C(5,5) = 1. Therefore, the required total is 10 + 5 + 1 = 16. Equivalently, subtract the subsets of sizes 0, 1, and 2: 32 − [C(5,0) + C(5,1) + C(5,2)] = 32 − (1 + 5 + 10) = 16.

At least 5 elements means that the subset may contain 5, 6, or 7 elements. The numbers of such subsets are C(7,5) = 21, C(7,6) = 7, and C(7,7) = 1. Therefore, the total is 21 + 7 + 1 = 29. The phrase 'at least' requires inclusion of every size greater than or equal to 5, so option B is correct.

The element 1 must be included in every required subset, so it is fixed. We only need to choose the remaining 2 elements from the other 5 elements, namely {2, 3, 4, 5, 6}. The number of choices is C(5,2) = 5 × 4 / 2 = 10. Thus, exactly 10 three-element subsets contain 1, making option B correct.

A 3-element subset that does not contain 1 must be formed entirely from the remaining five elements, {2, 3, 4, 5, 6}. Thus, the problem is equivalent to choosing 3 elements from these 5 elements. The number of choices is C(5,3) = 5!/(3!2!) = (5 × 4 × 3)/(3 × 2 × 1) = 10. Hence, exactly 10 three-element subsets of A exclude the element 1.

A member S of P(U) contains A as a subset when every element of A, namely a, c, and e, is already included in S. The remaining elements b, d, f, and g are optional; each may be included or excluded independently. Thus there are 2^4 = 16 possible choices for the optional elements. Hence 16 members of P(U) contain A, so option B is correct.

With respect to U, A' = {1, 3, 5, 7}. A subset S of U is disjoint from A' exactly when S contains none of 1, 3, 5, or 7. Therefore S can contain only elements of A = {2, 4, 6, 8}. Every subset of A is disjoint from A', and A has 4 elements, so the number is 2^4 = 16. Hence option B is correct.

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.

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.

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.

The set A contains four elements. Its complement relative to the given universal set U therefore contains 10 − 4 = 6 elements; explicitly, A′ = {1,4,6,8,9,10}. Every element may either be included or excluded from a subset, so a six-element set has 2^6 = 64 subsets. Consequently, |𝒫(A′)| = 64, and option C is the only correct answer.

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.

The elements of A are ∅, the number 1, and the set {1}. These are three distinct objects: 1 is not the same as {1}, and ∅ is different from both. Hence |A| = 3. The power set of a set with n elements has 2^n members, including the empty subset and the complete set. Therefore |𝒫(A)| = 2^3 = 8, so option C is correct.

A member of 𝒫(A) with exactly two elements is a two-element subset of the four-element set A. We must choose 2 elements from 4, and the order of selection does not matter. Therefore the number is the combination C(4,2) = 4!/(2!2!) = 6. The six subsets are {a,b}, {a,c}, {a,d}, {b,c}, {b,d}, and {c,d}. Hence option B is correct.