Power Set and Subsets

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.

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.

If a set A has 4 elements, then its power set P(A) has 2^4 = 16 elements. A singleton member of P(P(A)) is a one-element subset whose only element is one member of P(A). There is exactly one singleton subset for each element of P(A), so P(P(A)) contains 16 singleton members. Therefore, option C is correct.

The universal set U contains the integers 1 through 15, so it has 15 elements. The numbers divisible by 3 are 3, 6, 9, 12, and 15; hence |A| = 5. Its complement A' therefore contains 15 − 5 = 10 elements. A set with 10 elements has 2^10 subsets, so |P(A')| = 2^10. Therefore option C is correct.

Because the endpoints are included, the integers in A are −2, −1, 0, 1, 2, and 3. Thus A contains 6 elements. For every finite set with n elements, the number of subsets in its power set is 2^n, because each element can either be selected or not selected. Therefore |P(A)| = 2^6 = 64. Hence option C is correct.

An n-element set has 2^n total subsets. A proper subset is any subset other than the set itself, so the number of proper subsets is 2^n − 1. Given 2^n − 1 = 31, adding 1 gives 2^n = 32 = 2^5. Therefore n = 5. For comparison, n = 4 would give 15 proper subsets and n = 6 would give 63, so option B is the only correct answer.

For every finite set S, the number of elements in its power set is |P(S)| = 2^|S|. Since |P(A′)| = 1024 = 2^10, we obtain |A′| = 10. The complement A′ is taken relative to U, so A and A′ partition U and |A| + |A′| = |U|. Hence |A| = 18 − 10 = 8. Applying the power-set formula again gives |P(A)| = 2^8. Therefore option B is correct. Option C, 2^10, is the size of P(A′), not the size of P(A), which is the key distinction in this problem.

A set with n elements has 2^n subsets because each element has two choices: it may be included or excluded. Here |A| = 4, so P(A) contains 2^4 = 16 subsets. The set A itself is one of these subsets, so excluding A gives 16 - 1 = 15. Therefore, option B is correct.

An n-element set has equally many subsets of even and odd cardinality. Since A has 6 elements, its total number of subsets is 2^6 = 64. Exactly half have even cardinality and half have odd cardinality, so the number of even-cardinality subsets is 64/2 = 32. This includes the empty set, whose cardinality is zero and is even.

The elements 1 and 2 are compulsory, while 5 is forbidden. Thus only 3 and 4 remain optional. Each optional element can independently be included or excluded, giving 2 choices for each and therefore 2^2 = 4 valid subsets. They are {1,2}, {1,2,3}, {1,2,4}, and {1,2,3,4}. Hence option B is correct.

The total number of subsets of A is 2^5 = 32. It is easier to count the complement: subsets containing neither a nor b. Such subsets can use only c, d, and e, giving 2^3 = 8 subsets. Therefore, subsets containing at least one of a or b equal 32 - 8 = 24. Option C is correct.

Use De Morgan’s law or calculate directly. A ∪ B = {1,2,3,4,5,6}, so its complement in U is {7,8}; hence A' ∩ B' = {7,8}. This set has two elements, and the power-set rule gives |P(A' ∩ B')| = 2² = 4. Therefore option B is correct. The other values correspond to using an incorrect number of elements in the final set.

By De Morgan’s law, A' ∪ B' = (A ∩ B)'. The intersection A ∩ B is {4,5}, containing 2 elements. Its complement in the 9-element universal set U therefore contains 9 - 2 = 7 elements. The power set of a 7-element set has 2^7 = 128 members. Hence option D is correct.

The union is A ∪ B = {1,2,3,4,5,6}, so it has 6 elements and its power set has 2^6 = 64 elements. The intersection is A ∩ B = {3,4}, so it has 2 elements and its power set has 2^2 = 4 elements. Hence the required difference is 64 − 4 = 60, so option C is correct.

A and its complement A' are disjoint and together form the universal set U. Therefore |U| = |A| + |A'| = 4 + 5 = 9. The power set of an n-element set has 2ⁿ elements, so |P(U)| = 2⁹ = 512. Option B is correct. The other choices are 2⁸, 2¹⁰, and 2¹¹, which use an incorrect size for U.

Let the fixed 4-element subset be F. A subset of U is disjoint from F precisely when it contains none of the elements of F. Therefore, its elements can be selected only from the remaining 10 − 4 = 6 elements. Each of these six elements has two independent choices: included or excluded. Hence the number of such subsets is 2^6 = 64. This is a direct application of the rule that an n-element set has 2^n subsets, with the forbidden elements removed.

The common elements of A and B are 4 and 10, so A ∩ B = {4,10}. The power set contains every subset of this two-element set. These are the empty set, the two singleton subsets {4} and {10}, and the whole set {4,10}. Therefore P(A ∩ B) = {∅,{4},{10},{4,10}}, which is option C. The universal set U is not needed after finding the intersection.

The complement of A relative to U consists of the elements of U that are not in A. Therefore A' = U − A = {5,6}, so |A'| = 2. Every element of A' can independently be either included in or excluded from a subset. Consequently, the power set P(A') contains 2^2 = 4 subsets: ∅, {5}, {6}, and {5,6}. Thus the correct answer is option B.

The universal set contains the integers from 1 through 10. Removing the elements of A leaves A' = {1,3,5,7,9}, which has five elements. A singleton subset of A' contains exactly one element, so there is one singleton subset for each element of A': {1}, {3}, {5}, {7}, and {9}. Hence P(A') contains exactly five singleton sets, making option B correct. The total size of P(A') is 32, but that is not the number of singleton members.

A set is always an element of its own power set, so A ∈ P(A). If P(A) = P(B), then every subset of A is also a subset of B and vice versa. In particular, each element of A, viewed as a singleton subset, belongs to P(B), which implies that it belongs to B; thus A ⊆ B. By the same argument B ⊆ A. Therefore A = B, so option B is the only valid conclusion. Equal power sets cannot arise from different original sets.

First find the union: A ∪ B = {a, c, d}. The complement is taken relative to U, so (A ∪ B)' = U − {a, c, d} = {b, e}. This set has two elements. A set with n elements has 2ⁿ subsets in its power set; therefore, n(P((A ∪ B)')) = 2² = 4. Hence option B is correct.

Using De Morgan’s law, A' ∩ B' = (A ∪ B)'. The union A ∪ B is {1, 2, 3, 4, 5, 6}; therefore its complement in U is {7, 8}. Thus A' ∩ B' has two elements. The power set of any two-element set contains 2² = 4 subsets, including the empty set and the set itself. Therefore the correct answer is option B, 4.

An element of P(A) is a subset of A. To form a subset containing exactly three elements from the five elements of A, choose any 3 of them. The number of such choices is the combination 5C3 = 5!/(3!2!) = (5 × 4)/(2 × 1) = 10. Therefore, exactly 10 elements of P(A) have cardinality 3, so option B is correct.

The condition requires 1 to be included and 4 to be excluded. These two elements therefore have fixed statuses. The remaining elements 2 and 3 are unrestricted; each can be included or omitted independently. Consequently, the number of valid subsets is 2² = 4. They are {1}, {1, 2}, {1, 3}, and {1, 2, 3}. Hence option B is the only correct answer.