Power Set and Subsets

Correct answer: D. {∅,{∅}}

Explanation: The power set of the empty set contains exactly one subset, namely the empty set itself: P(∅) = {∅}. Now take the power set of the one-element set {∅}. Its subsets are the empty set and the set containing its only element, namely {∅}. Therefore P(P(A)) = P({∅}) = {∅,{∅}}. Notice that ∅ and {∅} are different: the first has no elements, while the second has one element. Hence option D is correct.

Correct answer: C. 32

Explanation: A six-element set has 2⁶ = 64 subsets in total. The subsets can be paired with their complements. Because the set has six elements, the complement of an even-sized subset is also even-sized, while the standard binomial identity shows that even- and odd-sized subsets occur equally often. Therefore half of 64, namely 2⁵ = 32, have even cardinality. Option C is correct.

Correct answer: C. 16

Explanation: For every non-empty finite set with n elements, the numbers of even-cardinality and odd-cardinality subsets are equal. Since A has five elements, P(A) has 2⁵ = 32 subsets in total. Therefore the number with an odd number of elements is half of 32, or 2⁴ = 16. Equivalently, 5C1 + 5C3 + 5C5 = 5 + 10 + 1 = 16. Thus option C is correct.

Correct answer: B. {∅, {∅}}

Explanation: The power set P(A) is the set of all subsets of A. Every set is a subset of itself, so A itself must be an element of P(A). Here A = {∅, {∅}}, where its two distinct elements are ∅ and {∅}. Therefore, the element {∅, {∅}} must occur in P(A), making option B correct.

Correct answer: C. 120

Explanation: If A has 4 elements, then its power set P(A) has 2⁴ = 16 elements. The question does not ask for the number of 2-element subsets of A; it asks for the number of 2-element subsets of P(A), which is a new set containing 16 elements. Therefore, we choose 2 elements from 16: C(16, 2) = 16 × 15 ÷ 2 = 120. Hence option C is correct. The key idea is to calculate the size of the power set first and then use combinations.

Correct answer: B. P(A ∩ B)

Explanation: An element of P(A) ∩ P(B) is a set that is a subset of both A and B. A set is a subset of both precisely when every one of its elements belongs to A ∩ B. Thus the common elements of the two power sets are exactly all subsets of A ∩ B, giving the identity P(A) ∩ P(B) = P(A ∩ B).

Correct answer: C. P(A) ∪ P(B) = P(A ∪ B)

Explanation: The identity P(A) ∪ P(B) = P(A ∪ B) is generally false. The left side contains sets that are subsets entirely of A or entirely of B, whereas the right side also contains mixed subsets containing elements from both A and B. For example, if A={1} and B={2}, {1,2} belongs to P(A∪B) but not to P(A)∪P(B).

Correct answer: B. 8

Explanation: For a finite set X, the number of elements in its power set is n(P(X)) = 2ⁿ⁽ˣ⁾. Since n(P(A)) = 16 = 2⁴, we get n(A) = 4. The complement therefore has n(A′) = n(U) − n(A) = 7 − 4 = 3 elements. Hence n(P(A′)) = 2³ = 8. Option B is correct. The value 16 belongs to P(A), while 4 is n(A), not the requested power-set cardinality.

Correct answer: C. A = A′

Explanation: For any sets X and Y, P(X) = P(Y) implies X = Y, because each set can be recovered as the union of all members of its power set. Therefore P(A′) = P(A) requires A′ = A. In an ordinary nonempty universal set, a set cannot equal its own complement, because A and A′ are disjoint; thus this equality is possible only in the degenerate case U = ∅. Among the given conditions, C states the required condition.

Correct answer: B. 57

Explanation: A six-element set has 2⁶ = 64 total subsets. The subsets with fewer than two elements are the empty subset, counted by C(6,0)=1, and the six one-element subsets, counted by C(6,1)=6. Thus 7 subsets have fewer than two elements. Subtracting them from all subsets gives 64 − 7 = 57 subsets containing at least two elements. Therefore option B is correct.

Correct answer: B. 8

Explanation: Only the odd elements 1 and 3 affect whether a subset sum is odd or even; the even elements 2 and 4 do not change parity. The sum is even when the subset contains an even number of odd elements: either neither 1 nor 3, or both 1 and 3. That gives 2 choices for the odd elements. The even elements are independently optional, giving 2² choices. Thus the total is 2 × 4 = 8, so option B is correct.

Correct answer: A. Only {1} ∈ P(A) is correct

Explanation: 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.

Correct answer: B. {{2}}

Explanation: 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.

Correct answer: B. 16

Explanation: 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.

Correct answer: C. 32

Explanation: 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.

Correct answer: C. 11

Explanation: 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.

Correct answer: C. 4

Explanation: 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.