Power Set and Subsets

The complement of A with respect to U is A′ = U − A = {2,4,6,8}. It has four elements, and every one of them is even. A set with n elements has 2ⁿ subsets, so P(A′) contains 2⁴ = 16 subsets. Since A′ itself contains only even numbers, every subset of A′ also contains only even numbers, including the empty set. Therefore, the correct answer is 16.

The power set P(A) consists of all subsets of A. Because {1} is a subset of A, it is an element of P(A), so {1} ∈ P(A) is true. For {1} ⊆ P(A) to be true, the element 1 itself would have to belong to P(A). However, P(A) contains sets such as ∅, {1}, and {1,2}, not the number 1 as an individual element. Hence only the first statement is correct.

The set A has exactly two elements: the number 1 and the set {2}. An element of P(A) must be a subset of A, so all of its elements must be either 1 or {2}. The set {{2}} has the single element {2}, which belongs to A; therefore {{2}} ⊆ A and {{2}} ∈ P(A). In contrast, {2} contains the number 2, which is not an element of A, and {1,2} also contains 2. Thus option B is correct.

A has three elements, so its power set has 2³ = 8 elements. The power set includes both the empty set ∅ and the original set A. The expression P(A) − {∅, A} removes exactly these two distinct elements from the power set. Therefore, the remaining number of elements is 8 − 2 = 6. These remaining sets are precisely the non-empty proper subsets of A, so option C is correct.

First find the intersection: A ∩ B = {2}. The complement is taken relative to the stated universal set U, so (A ∩ B)′ = U − {2} = {1,3,4,5}. This complement has four elements. The power set of a four-element set has 2⁴ = 16 elements. Therefore n(P((A ∩ B)′)) = 16. The value 4 is only the size of the complement, not the size of its power set.

Using De Morgan’s law, A′ ∪ B′ = (A ∩ B)′. The intersection A ∩ B is {3}, so its complement in U is {1,2,4,5,6}, which has five elements. The power set of a five-element set contains 2⁵ = 32 elements. Therefore n(P(A′ ∪ B′)) = 32. Directly, A′ = {4,5,6} and B′ = {1,2,6}; their union is also {1,2,4,5,6}, confirming the result.

For a set A with three elements, n(P(A)) = 2³ = 8. Applying the power-set operation again gives n(P(P(A))) = 2⁸ = 256. Applying it a third time gives n(P(P(P(A)))) = 2²⁵⁶. The exponent is therefore 256, not 16 or 8. Option D is only the size of P(P(A)), so it stops one power-set operation too early. Hence option C is correct.

There are 2³ = 8 subsets of A that contain 1, because the other three elements may be chosen freely. There are C(4,2) = 6 two-element subsets. Some subsets belong to both groups: a two-element subset containing 1 is formed by choosing one of the remaining three elements, giving C(3,1) = 3. By inclusion–exclusion, the required count is 8 + 6 − 3 = 11. Therefore option C is correct.

Any required subset must exclude a, so only the three-element set {b,c,d} is available. We need its subsets of odd size. Among a three-element set, the odd-sized subsets consist of the three one-element subsets and the one three-element subset, for a total of 3 + 1 = 4. Equivalently, half of the 2³ = 8 subsets have odd cardinality, giving 2² = 4. Hence option C is correct.

The element 1 is forbidden, while 2 is compulsory. After fixing these conditions, the remaining elements 3, 4, and 5 may each be either included or excluded independently. Thus there are 2 choices for each of three unrestricted elements, giving 2³ = 8 valid subsets. The compulsory element 2 does not add a choice, and the forbidden element 1 has no choice. Therefore option B is correct.

A set with n elements has 2^n subsets, so its power set contains 2^n elements. Therefore, n(P(B)) = 2^5 = 32 and n(P(A)) = 2^3 = 8. Since A is a subset of B, every subset of A is also a subset of B; hence P(A) is a subset of P(B). Thus, P(B) − P(A) contains 32 − 8 = 24 elements. Therefore, option C is correct.

The complement is taken relative to U. Thus A′ = U − A = {1, 3, 5, 7}. We need subsets of A′ that must contain 1 and 7. Those two elements are fixed as included. The remaining elements, 3 and 5, can independently be included or excluded, giving 2 choices for each. Therefore the number of valid subsets is 2² = 4. Option B is correct; 16 counts all subsets of A′ without the required-element condition.

An element of P(A) is itself a subset X of A. Since X ⊆ A, we have X ∩ A = X. For this intersection to be empty, X must be the empty set ∅. Thus the only member of P(A) disjoint from A is ∅, so the number is 1. Although P(A) has 2⁴ = 16 members, every non-empty subset shares at least one element with A. Therefore option B is correct.

The even elements of U form A = {2, 4, 6, 8, 10}. Therefore its complement is A′ = {1, 3, 5, 7, 9}, which has 5 elements. The elements of P(A′) are all subsets of A′. A subset with exactly two elements can be selected in C(5, 2) ways = 5!/(2!3!) = 10. Hence P(A′) contains 10 two-element subsets, so option B is correct.

The set A' ∩ B consists of elements that are in B but not in A. Set B is partitioned into two disjoint parts: A ∩ B and A' ∩ B. Therefore n(B) = n(A ∩ B) + n(A' ∩ B). Substituting the given values gives 53 = 22 + n(A' ∩ B), so n(A' ∩ B) = 53 − 22 = 31. Thus option A is correct.

For any two finite sets, \(n(A\cup B)=n(A)+n(B)-n(A\cap B)\). Since \(A\cap B=\varnothing\), the sets are disjoint and their intersection has zero elements. Therefore, \(n(A\cup B)=3+5-0=8\). Option B counts only set B, option C counts only set A, and option D incorrectly multiplies the two cardinalities. Thus option A is correct.

The only common element of A and B is 3, so \(A\cap B=\{3\}\). This intersection has one element. The power set of an n-element set contains \(2^n\) subsets, including the empty set and the original set. Therefore \(|\mathcal{P}(A\cap B)|=2^1=2\), making option A correct.

Rewrite the condition as y = 1 - x and test the elements of A. For x = -1, y = 2, giving (-1,2); for x = 0, y = 1, giving (0,1); for x = 1, y = 0, giving (1,0); and for x = 2, y = -1, which is not in B. Thus exactly three ordered pairs belong to A × B and satisfy the equation. Therefore, option B is correct.