Power Set and Subsets

The power set P(A) contains every subset of A. If the element 1 must be included, then the remaining elements 2 and 3 can each be either included or excluded independently. Thus there are 2 choices for 2 and 2 choices for 3, giving 2×2=2^2=4 subsets: {1}, {1,2}, {1,3}, and {1,2,3}. Therefore, option A is correct.

A subset that does not contain p can be formed only from the other three elements q, r, and s. Each of these three elements has two independent possibilities: it may be included or excluded. Therefore the number of permissible subsets is 2^3=8. These include the empty set, all single-element choices, all two-element choices, and the set {q,r,s}. Hence option A is correct.

The universal set U contains 15 elements, while A contains the five multiples of 3 from 1 to 15. Therefore, the complement A' contains the remaining 15 − 5 = 10 elements. A set with n elements has 2^n subsets, so its power set contains 2^10 = 1024 elements. Hence, n(P(A')) = 1024, which is option A.

The notation distinguishes an element from a subset. Here the set {1,2} is explicitly listed as one element of A, so {1,2}∈A. It is also a subset of A because both 1 and 2 are individually elements of A. Every subset of A belongs to P(A), so {1,2}∈P(A) as well. Therefore option A is the only correct statement.

A subset with at least four elements must have either exactly four elements or exactly five elements, since A itself has only five elements. The number of four-element subsets is C(5,4)=5, and the number of five-element subsets is C(5,5)=1. Adding them gives 5+1=6. Therefore P(A) has exactly six subsets with at least four elements, making option A correct.

For a three-element set, an odd-sized subset has either one element or three elements. There are C(3,1)=3 one-element subsets: {a}, {b}, and {c}. There is C(3,3)=1 three-element subset, namely {a,b,c}. Thus the total number is 3+1=4, so option A is correct. The empty set and two-element subsets have even size and are excluded.

The statement A⊆B means every element of A is also an element of B. Take any set X in P(A). By definition, X⊆A. Since every element of A belongs to B, all elements of X also belong to B, so X⊆B. Therefore X∈P(B). This proves P(A)⊆P(B), making option A always true. Equality occurs only in special cases, such as A=B.

The condition fixes two elements: 1 must be included and 4 must be excluded. The remaining elements 2 and 3 are unrestricted, so each can independently be selected or not selected. Hence the number of valid subsets is 2^2=4. Explicitly, they are {1}, {1,2}, {1,3}, and {1,2,3}. Thus option A is the only correct answer.

Let S be a subset of U. If its complement U − S is a singleton, then exactly one element of U is outside S, so S must contain exactly three elements. The number of three-element subsets of a four-element set is C(4,3) = 4. Therefore, exactly four elements of P(U) have singleton complements.

The empty set has no elements, so its only possible subset is the empty set itself. Therefore, when \(A=\varnothing\), its power set is \(\mathcal{P}(A)=\{\varnothing\}\). Notice the distinction: \(\varnothing\) is the sole element of the power set, whereas \(\{\varnothing\}\) is a set containing one element. Hence the power set has exactly one element, not zero.

The number of two-element subsets of a five-element set is C(5,2) = 10. The number of three-element subsets is C(5,3) = 10. These two classes are disjoint because a subset cannot simultaneously have two and three elements. Therefore, the required total is 10 + 10 = 20.

Although the word “sets” has five positions, repeated letters are counted only once in a set. Its distinct letters are {s, e, t}, so n(A) = 3. A set with n elements has 2ⁿ subsets; consequently, n(P(A)) = 2³ = 8. Hence option A is the correct answer.

Solve the defining equation: \(x^2=x\) implies \(x^2-x=0\), or \(x(x-1)=0\). Hence \(x=0\) or \(x=1\), so \(A=\{0,1\}\). Its complement in \(U\) is \(A'=\{2,3,4,5\}\), which has four elements. A set with \(k\) elements has \(2^k\) subsets, so \(n(\mathcal{P}(A'))=2^4=16\). Therefore option A is correct; 4 is only the size of \(A'\), not of its power set.

The elements a and b must be included in every required subset, so their choices are fixed. The remaining elements c, d, and e are each optional: each can either be included or excluded. Thus there are 2 choices for each of three independent elements, giving 2³ = 8 subsets. Therefore, option A is correct.

The perfect squares in the universal set \(U=\{1,2,\ldots,25\}\) are \(1,4,9,16,25\), because they are \(1^2,2^2,3^2,4^2,5^2\), respectively. Thus \(A\) contains five elements. Since \(U\) contains 25 elements and \(A'\) consists of all elements of \(U\) that are not perfect squares, \(n(A')=n(U)-n(A)=25-5=20\). Therefore option A is the unique correct answer.

A has three elements, so its power set contains 2³ = 8 subsets in total. A proper subset is a subset that is not equal to the original set A. Every subset in P(A) is contained in A, but A itself must be excluded. Therefore, the number of proper subsets is 8 − 1 = 7.

Since 2 must be included, the subset already has one element. To make its total size even, we must choose an odd number of elements from the remaining set {1, 3, 4}. We can choose one element in C(3,1)=3 ways or all three elements in C(3,3)=1 way. Thus the total is 3+1=4.

P(A) consists of all subsets of A: ∅, {a}, {b}, {c}, {a,b}, {a,c}, {b,c}, and {a,b,c}. A subset is disjoint from A only when it has no element in common with A. Since every non-empty subset listed contains elements of A, only the empty set qualifies. Hence the number is 1, option A.

A set with six elements has 2^6 = 64 subsets. To count subsets containing 1 or 2, use the complement: subsets containing neither 1 nor 2 can choose freely only from {3,4,5,6}, giving 2^4 = 16. Therefore the required number is 64 - 16 = 48. Option A is correct.

The complement A' is defined relative to U, so A' = {3,4,5}. Every element of A' is therefore an element of U, which means A' ⊆ U. Whenever one set is a subset of another, every subset of the smaller set is also a subset of the larger set. Consequently P(A') ⊆ P(U), making option A correct.