Power Set and Subsets

The governing definition is P(A) = the set of all subsets of A. Here A has two elements: ∅ and {∅}. The set {{∅}} contains the single element {∅}, which is an element of A; hence {{∅}} is a subset of A and therefore belongs to P(A). Option A is correct. Option B is a membership statement rather than the required subset, and 1 and {1} contain objects unrelated to A, so they are not subsets of A.

The set A has exactly two elements: ∅ and {∅}. Its subsets are ∅, {∅}, {{∅}}, and {∅, {∅}}. Option D is {∅, {{∅}}}; it contains the element {{∅}}, but {{∅}} is not an element of A. A contains {∅}, not {{∅}}. Therefore every element of option D is not in A, so option D is not a subset. Careful counting of nested braces is essential in this question.

A has three elements: the number 1, the number 2, and the set {3, 4}. A set C is a subset of A when every element of C is also an element of A. In option B, the elements are 2 and {3, 4}; both occur directly in A, so B is a subset of A. In option A, 3 is not an element of A; only the set {3, 4} is present. Similarly, 4 and {3} are not direct elements of A in option D.

The elements of A are the empty set ∅, the singleton set {∅}, and the number 1. A set is a subset of A only when every one of its elements is an element of A. Option A contains ∅ and 1, both of which belong to A. Option B contains 0, option C contains {{∅}}, and option D contains 2; these are not elements of A.

The elements of A are the set {1,2}, the number 3, and the number 4. A subset may contain only elements that are directly in A. Options A, B, and C use 3, 4, and/or the whole element {1,2}, so they are subsets. Option D contains 1 and 2 separately, but neither is a direct element of A; therefore it is not a subset.

Each member X of A must contain both 1 and 4 because {1,4} ⊆ X. The element 2 is forbidden, so it cannot occur in X. The only remaining element from the universal set is 3, and it can be either omitted or included. Consequently, the only possibilities are X = {1,4} and X = {1,3,4}. Thus A = {{1,4}, {1,3,4}}, making option A correct. The other choices either include the forbidden element 2 or omit a required element.

The power set P(A) consists of all subsets of A. The empty set ∅ is a subset of every set, and both {∅} and {1} are also subsets of A. However, the expression {1, {∅}} contains two elements, 1 and {∅}, and both are elements of A; therefore it is a subset of A and hence belongs to P(A). It is not itself an element of A, because the elements listed in A are ∅, {∅}, and 1. Thus option D is correct.

If A has 3 elements, then its power set P(A) contains 2³ = 8 elements. A set containing 8 elements has 2⁸ = 256 subsets in total. Proper subsets are all subsets except the set itself, so we must remove one subset, namely P(A) itself. Thus the number of proper subsets is 256 − 1 = 255. Therefore option B is correct. The important point is to first find the size of P(A), and only then count its subsets.

The power set P(A) consists of all subsets of A. The set A has two elements: the number 0 and the set {0}. The set {0} contains only the element 0, which belongs to A, so {0} is a subset of A and therefore an element of P(A). In contrast, 0 is an element of A but is not itself a subset of A.

The elements of A are ∅, {1}, and 2. An element of P(A) must be a subset whose every member is one of these three objects. The set {∅, 2} uses two actual elements of A, so it is a subset of A and therefore belongs to P(A). In contrast, 1 is not an element of A, and neither {1,2} nor {{2}} satisfies the subset condition. Thus A is correct.

The elements of A are 1 and the set {2}; the number 2 itself is not an element of A. Every subset of A belongs to P(A). Since {2} is an element of A, the singleton set whose only element is {2}, written {{2}}, is a subset of A and therefore belongs to P(A). Option A is an element of A, not necessarily a subset, while option C is not even an element of A.

For an empty set A = ∅, its power set contains exactly one subset: P(A) = {∅}. This new set has one element, namely ∅. The power set of a one-element set has 2¹ = 2 elements: the empty subset and the set containing that one element. Thus P(P(A)) = {∅, {∅}}, so it contains two elements. The answer is not 1 because the second power-set operation is also required.

The elements of P(A) are all subsets of A. We need to list subsets whose elements add to 3: {3}, {0, 3}, {1, 2}, and {0, 1, 2}. These are four different subsets. The zero element is important because adding 0 does not change a subset’s sum, so it creates an additional valid subset from both {3} and {1, 2}. No other subset has sum 3. Therefore, the correct answer is option C, 4.

For A = {m, n}, its power set is P(A) = {∅, {m}, {n}, {m, n}}. A set X is a subset of P(A) only when every element of X is itself an element of P(A). In option C, the elements are m and n, but m and n are not subsets of A and therefore are not elements of P(A). The other choices contain only valid members of P(A), or are empty. Thus option C is the only set that is not a subset of P(A).

A set with n elements has exactly 2^n subsets, so |P(A)| = 2^n = 64 = 2^6. Hence n = 6. The number of 3-element subsets of a six-element set is the combination C(6, 3) = 6!/(3!3!) = (6 × 5 × 4)/(3 × 2 × 1) = 20. Therefore, option C is correct. The value 15 would count 2-element subsets, not 3-element subsets, which is a common error.

The wording asks for subsets of \(A\) having even cardinality; these are the elements of the power set \(\mathcal{P}(A)\). Since \(|A|=5\), the number of even-cardinality subsets is \(\binom50+\binom52+\binom54=1+10+5=16\). Equivalently, for every nonempty set, even- and odd-cardinality subsets occur equally often, so each group has \(2^{5-1}=16\) subsets. Thus option B is correct.

We count subsets of A that must contain a and must not contain both b and c. The element a is fixed as included. For b and c, the allowed choices are neither, b only, or c only, giving 3 choices; the choice containing both is excluded. Elements d and e are independent, with 2 choices each. Therefore the total is 1 × 3 × 2 × 2 = 12.

A subset of size 2 can be selected from a six-element set in C(6,2) ways, and a subset of size 4 can be selected in C(6,4) ways. These two classes do not overlap because a subset cannot have both sizes simultaneously. Thus the required number is C(6,2) + C(6,4) = 15 + 15 = 30.

The elements of A are ∅, {a}, and b. A member of P(A) must be a set whose every element belongs to A. In option A, the elements are ∅ and {a}; both are elements of A, so {∅,{a}} is a subset of A and therefore belongs to P(A). The other choices contain elements not in A or are not subsets.

The statement |P(A)| = 256 means that P(A) itself is a set containing 256 elements. The power set of any finite set with n elements contains 2ⁿ elements, since every element has two choices: included or excluded. Therefore |P(P(A))| = 2²⁵⁶, not 256² or 2⁸.

A has 7 elements, so its power set P(A) contains 2^7 = 128 subsets in total. The number of subsets having exactly 5 elements is the binomial coefficient C(7,5) = 21, because we choose 5 elements from the 7 elements of A. The required number is the total number of subsets minus these subsets: 128 − 21 = 107. Therefore option B is correct.

There are 15 numbers from 1 to 30 divisible by 2 and 10 divisible by 3. Numbers divisible by both 2 and 3, that is, divisible by 6, are counted twice; there are 5 of them. By inclusion–exclusion, |A| = 15 + 10 − 5 = 20. Since U has 30 elements, the complement has |A′| = 30 − 20 = 10 elements. Hence option A is correct.

For any finite set X, |P(X)| = 2^|X|. From |P(A)| = 32 = 2^5, we get |A| = 5. Similarly, |P(B)| = 8 = 2^3, so |B| = 3. The Cartesian product A × B therefore has |A||B| = 5 × 3 = 15 ordered pairs. Its power set consequently has 2^15 elements. Hence option A is correct.

There are 2^5 = 32 subsets in P(A). Pair every subset S with S △ {1}, where 1 is odd; toggling 1 changes the parity of the subset sum. Thus each pair contains one even-sum subset and one odd-sum subset. Consequently the 32 subsets divide equally, and the number with an even sum is 32/2 = 16.

The power set P(A) contains every subset of A. The elements of A are 1, the set {2}, and 3; importantly, 2 itself is not an element of A. The set {1, {2}} uses the elements 1 and {2}, both of which belong to A, so it is a subset of A and therefore belongs to P(A). The other choices do not satisfy this condition, so option A is correct.