Power Set and Subsets

A∩B=∅ means A and B have no common elements. Any set belonging to both P(A) and P(B) must be a subset of both A and B, so it must be a subset of A∩B. The only subset of the empty set is ∅. Hence P(A)∩P(B)={∅}, whose cardinality is 1, so option B is correct.

A set belongs to both P(A) and P(B) exactly when it is a subset of both A and B. Such sets are precisely the subsets of A∩B, so P(A)∩P(B)=P(A∩B). Since |A∩B|=2, its power set has 2^2=4 members. Therefore option B is correct; the separate sizes of A and B are not needed.

Since A and B are disjoint and each has two elements, A ∪ B = {1, 2, 3, 4}, so |P(A ∪ B)| = 2^4 = 16. Also, |P(A)| = |P(B)| = 2^2 = 4. The two power sets share only the empty set, because no nonempty subset can consist simultaneously of elements from A and from B. Thus |P(A) ∪ P(B)| = 4 + 4 − 1 = 7, and the required difference has 16 − 7 = 9 members.

The set A has 3 distinct elements. A two-element subset is formed by choosing any 2 of these 3 elements, so the required number is C(3, 2) = 3!/(2!1!) = 3. Explicitly, the subsets are {p, q}, {p, r}, and {q, r}. Each of these is a member of P(A), because the power set contains every subset of A. Therefore, option B is correct.

The power set P(A) consists of all subsets of A. A member of P(A) having exactly 4 elements is therefore a 4-element subset of A. Since A has 6 elements, the number of such subsets is C(6, 4) = 6!/(4!2!) = (6 × 5)/(2 × 1) = 15. Equivalently, choosing 4 elements to include is the same as choosing the 2 elements to exclude, so C(6, 4) = C(6, 2) = 15. Hence option B is correct.

For a set with n elements, exactly half of its 2^n subsets have even cardinality and half have odd cardinality. Therefore, the number of odd-cardinality subsets is 2^(n−1). With n = 7, the count is 2^6 = 64. This includes all possible odd sizes, namely 1, 3, 5, and 7. Option B is correct.

A set with 8 elements has 2^8 = 256 total subsets. For every nonempty finite set, the subsets with even cardinality and the subsets with odd cardinality occur in equal numbers. Thus the number of even-cardinality subsets is 2^8/2 = 2^7 = 128. The empty set is included because its cardinality is zero, which is even. Therefore, option B is correct.

Element 1 is compulsory, so it has only one status: it must be included. Element 5 is forbidden, so it also has only one status: it must be excluded. The remaining elements 2, 3, and 4 may each be included or excluded independently, giving 2^3 = 8 possible subsets. Hence option B is correct.

The elements a and b are required, so they are fixed as included in every valid subset. The remaining four elements, c, d, e, and f, are unrestricted. Each of these four elements has two independent choices: include it or exclude it. Therefore, the number of valid subsets is 2^4 = 16, so option B is correct.

There are 2^6 = 64 total subsets of A. It is easier to count the complement: subsets containing neither 2 nor 4 can use only the remaining four elements, so there are 2^4 = 16 of them. Therefore, subsets containing at least one of 2 or 4 equal 64 − 16 = 48. Option C is correct.

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.

First form the union: A ∪ B = {1, 2, 3, 5, 7, 8}. Taking the complement relative to the universal set U leaves the elements of U that are not in this union: (A ∪ B)' = {4, 6}. This complement has 2 elements. A set with n elements has 2^n subsets, so the power set P((A ∪ B)') has 2^2 = 4 members. Therefore, option B is correct. The order matters: find the union first, then take its complement in U, and finally count the power set.

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.

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.

To form a subset of A that does not contain a, we may choose or reject each of the remaining elements b, c, and d independently. Each element gives 2 choices, so the total number of permitted subsets is 2 × 2 × 2 = 2³ = 8. These include the empty set and all subsets formed only from {b,c,d}. Therefore, 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.

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.

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.

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.