Power Set and Subsets

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.

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.

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.

Set A contains two elements. For any finite set with n elements, its power set contains 2^n subsets because each element can either be included or excluded from a subset. Therefore, n(B) = n(P(A)) = 2^2 = 4. Explicitly, B = {∅, {1}, {2}, {1,2}}. The value 2 is n(A), not n(P(A)), while 8 would require a three-element original set.

The power set contains every subset of the original set, including the empty subset, both singleton subsets, and the original two-element set. Therefore, P({m,n}) = {∅, {m}, {n}, {m,n}}. The braces are important: m and n alone are elements, whereas {m} and {n} are singleton subsets. Thus option B lists all four subsets correctly.

To form a three-element subset from the five elements of A, choose any 3 elements without considering their order. Therefore, the required number is C(5,3) = 5!/(3!2!) = (5×4×3)/(3×2×1) = 10. These are ten distinct members of P(A). The total number of all subsets is 2^5 = 32, but only the subsets with exactly three elements are counted here.

For a finite set with n elements, the number of members of its power set is 2ⁿ. Each of the six elements of A can either be selected or not selected when forming a subset, producing 2 × 2 × 2 × 2 × 2 × 2 = 2⁶ possibilities. Since 2⁶ = 64, the power set P(A) contains 64 subsets. Therefore option D is correct; 32 is only 2⁵.

A singleton subset contains exactly one element. The odd elements of A are 1 and 3, so the singleton subsets containing an odd number are {1} and {3}. There are exactly two such subsets. The even elements 2 and 4 produce {2} and {4}, but these do not satisfy the condition. Therefore option B is correct. In general, each qualifying element produces one distinct singleton subset.

A set with n elements has 2^n total subsets because each element can either be included or excluded. Here, |A| = 3, so P(A) has 2^3 = 8 subsets. Exactly one of them is the empty set, so the number of non-empty subsets is 8 − 1 = 7. Therefore, option B is correct; 8 counts the empty set too.

If a is not allowed in a subset, only b and c may be selected. Each of these two elements has two independent choices: include it or leave it out. Thus the number of permitted subsets is 2^2 = 4. They are the empty set, {b}, {c}, and {b, c}. Therefore, option C is correct.

The power set P(A) is the collection of all subsets of A. If a set has n elements, each element has two choices—being included or excluded—so the power set has 2^n elements. Here n = 4, giving |P(A)| = 2^4 = 16. Therefore, option C is correct. The value 4 is the size of A itself, not the size of its power set.

The complement A′ contains all elements of the universal set U that are absent from A. The set A contains the even numbers from 2 through 10, so removing them from U leaves the odd numbers 1, 3, 5, 7, and 9. Thus A′ = {1, 3, 5, 7, 9}, making option A correct. Option B is A itself, not its complement.

If a finite set A has n elements, then its power set 𝒫(A) has 2ⁿ elements. Here, 2ⁿ = 32 = 2⁵, so A has 5 elements. Therefore, A has 32 subsets in total. A proper subset is any subset other than A itself, so we exclude exactly one subset: A. Hence, the number of proper subsets is 32 − 1 = 31.

The power set of a set is the set containing all its subsets. The empty set has exactly one subset: the empty set itself. Therefore, 𝒫(∅) = {∅}. Notice the distinction between ∅, which has no elements, and {∅}, which has one element—the empty set. This distinction is essential when determining power sets and their cardinalities.

The set A contains five distinct elements: 2, 4, 6, 8, and 10. For every element, a subset can either include it or exclude it, giving two choices independently. Therefore, a set with n elements has 2ⁿ subsets. Here n = 5, so |𝒫(A)| = 2⁵ = 32. The requested number is the cardinality of the power set, not the number of elements in A.

The value n(𝒫(B)) = 128 tells us that B has 128 subsets in total. Every set is a subset of itself, but it is not a proper subset of itself. Therefore, exactly one subset—the set B—is removed when counting proper subsets. The number of proper subsets is 128 − 1 = 127. Although 128 = 2⁷ also shows that B has seven elements, that value is not the requested answer.