Power Set and Subsets

The condition requires 2 to be included and 3 to be excluded, so those two choices are fixed. Only 1 and 4 remain freely selectable. Each can be included or omitted independently, giving 2 × 2 = 2^2 = 4 valid subsets: {2}, {1,2}, {2,4}, and {1,2,4}. Therefore, option B is correct.

The three listed objects are distinct elements: ∅ is the empty set, {∅} is a one-element set whose element is ∅, and 0 is a number. Since A has 3 distinct elements, the number of its subsets is 2^3 = 8. Therefore n(P(A)) = 8. It is important not to confuse ∅ with {∅}; they are different sets.

The set A has 4 elements. A set with n elements has 2^n total subsets because each element can either be included or excluded. Therefore, A has 2^4 = 16 subsets. The number of subsets containing exactly two elements is given by the combination 4C2 = 4!/(2!2!) = 6. Hence, the required numbers are 16 and 6, so option B is correct.

The set A has three elements, so its power set contains 2^3 = 8 subsets. To count subsets containing at least one of a or b, subtract the subsets that contain neither a nor b. If neither a nor b is selected, only c may be selected or omitted, giving two subsets: the empty set and {c}. Thus, the required number is 8 - 2 = 6. Therefore, option C is correct.

The set A contains three elements. For any finite set with n elements, the power set contains 2^n elements, because every element has two choices: it may be included in a subset or excluded. Thus, |P(A)| = 2^3 = 8. Therefore, the power set has eight elements, and option B is correct. The value 3 is the number of elements of A, not of its power set.

The power set P(A) is the set of all subsets of A. For A = {a, b}, its elements are ∅, {a}, {b}, and {a, b}. The symbol a is an element of A, but it is not itself a subset of A; it is not enclosed in braces as a one-element set. Therefore, a does not belong to P(A), making option D correct. The distinction between a and {a} is essential.

The complement A' consists of all elements in the universal set U that are not present in A. Starting with U = {1, 2, 3, 4, 5}, remove the elements 2 and 4 belonging to A. The remaining elements are 1, 3, and 5. Thus A' = {1, 3, 5}, so option A is correct. The complement always depends on the specified universal set.

A set with four elements has 2^4 = 16 total subsets. A proper subset is a subset that is not equal to the original set A itself. Since exactly one of the 16 subsets is A, subtract it from the total: 16 − 1 = 15. The empty set is included among the proper subsets. Therefore, A has 15 proper subsets, so option B is correct.

If the original set A has n elements, then its power set has 2^n elements. The question gives 2^n = 32. Since 32 = 2^5, it follows that n = 5. Therefore, the original set A contains five elements. Choosing 4 would give only 16 subsets, while choosing 6 would give 64 subsets. Hence option B is correct.

Although A contains the empty set, A itself is not empty. The set A = {∅} has exactly one element: the empty set. A set with one element has 2^1 = 2 subsets. They are ∅ and {∅}. Therefore, P(A) = {∅, {∅}} and contains two elements. Confusing ∅ with {∅} would incorrectly lead to one, so option B is correct.

The power set P(A) is the set of all subsets of A. The set {p,q} is a subset of A because both of its elements belong to A. The set {p,q,r} is A itself, and every set is a subset of itself. Therefore both {p,q} and {p,q,r} belong to P(A), so both are elements of the power set. Option A is correct.

For A = {1,2}, the power set is P(A) = {∅, {1}, {2}, {1,2}}. A member of the power set must itself be a subset of A. The singleton set {1} is such a subset, so it is an element of P(A). The expression 1 is a number, not the set {1}; similarly, 2 and 3 are not suitable power-set elements in the listed form. Thus option B is correct.

A set containing n distinct elements has exactly 2ⁿ subsets, because each element can either be included in or excluded from a subset. Here A = {k,l,m} has three distinct elements. Therefore n(P(A)) = 2³ = 8. The power set includes the empty set, the three singleton subsets, the three two-element subsets, and A itself. Hence option C is correct.

The set A has exactly two elements: the number 1 and the set {2}. The inner set {2} counts as one element of A; the number 2 is not separately an element of A. A set with n elements has 2ⁿ subsets, so here the number is 2² = 4. Explicitly, the subsets are ∅, {1}, {{2}}, and {1,{2}}. Hence option C is correct.

First find the complement of A in U: A′ = U − A = {b,c}. Thus A′ has two elements. The power set of any finite set containing n elements has 2ⁿ elements, because every element has two choices—present or absent—in a subset. Therefore n(P(A′)) = 2² = 4. Option A gives the size of A′, not the size of its power set, so option B is correct.

The empty set is a subset of every set, including A = {2,4,6}. Since P(A) is the set of all subsets of A, the empty set must be one of its elements. Therefore ∅ ∈ P(A). It is important to distinguish the statements ∅ ⊆ A and ∅ ∈ P(A): the first says it is a subset of A, while the second says it is an element of the power set. Option A is correct.

The complement A′ consists of all elements of the universal set U that are not in A. Since A is the empty set, it contains no elements at all, so every element of U is outside A. Consequently, A′ = U = {1,2,3,4}. This illustrates the standard identity ∅′ = U, provided the complement is taken with respect to U. Therefore option B is correct.

A singleton subset contains exactly one element. For every one of the 5 elements of A, we can form one singleton subset, such as {a}, where a belongs to A. Therefore, the number of singleton subsets is equal to n(A), which is 5. Although P(A) has 2^5 = 32 subsets in total, only 5 of them contain exactly one element.

A two-element subset is obtained by choosing 2 different elements from the 4 elements of A. Since the order of selection does not matter, combinations are used: C(4,2) = 4!/(2!2!) = 6. Thus, P(A) contains exactly 6 subsets having two elements. The total power set has 2^4 = 16 subsets, but the question asks only for those of size two.

A three-element subset of A must contain all three elements because A itself has exactly three elements. Therefore, the only three-element subset is {1, 2, 3}, which is A itself. Since every subset of A is an element of P(A), this one subset belongs to P(A). Hence, the number of three-element subsets is C(3,3) = 1.