Power Set and Subsets

Any subset of A that contains {1, 2} must include both 1 and 2. The remaining elements 3 and 4 are optional, and each can independently be included or excluded. Thus the possible subsets are {1,2}, {1,2,3}, {1,2,4}, and {1,2,3,4}, giving 2^2 = 4 subsets. Therefore, option B is correct.

The set A contains the even numbers {2,4,6,8,10}. Therefore, its complement in U is A′ = {1,3,5,7,9}, which has 5 elements. A set with 5 elements has 2⁵ = 32 subsets in its power set. One of these subsets is the empty set, so the number of non-empty members is 32 − 1 = 31. Hence, option B is correct.

First, P(A) = {∅, {1}, {2}, {1,2}}. The statement {1} ⊆ P(A) means that every element of the set {1}, namely the object 1, must be an element of P(A). However, P(A) contains sets such as {1}, not the number 1 itself. Thus 1 ∉ P(A), so the statement is false. Option B is correct.

For the two elements 2 and 3, the condition allows exactly two possibilities: both are selected, or neither is selected. The other three elements, 1, 4, and 5, can each be selected or omitted independently, giving 2³ = 8 choices. Multiplying by the 2 allowed cases for the pair gives 2 × 8 = 16 subsets. Hence, option B is correct.

First find the intersection: A ∩ B = {3}, so it contains one element. Taking the complement relative to U gives (A ∩ B)' = U − {3} = {1, 2, 4, 5, 6}, which has 5 elements. A set with n elements has 2^n subsets, including the empty set and the set itself. Therefore, |P((A ∩ B)')| = 2^5 = 32, so option B is correct.

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.

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.

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.

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.

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.

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.

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.

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.

“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.