Power Set and Subsets

In set theory, repeated elements do not create new elements. Thus the given collection is the set A = {1,2,3}, whose cardinality is 3 rather than 5. A set with n elements has 2^n subsets, so n(P(A)) = 2^3 = 8. The repeated 1s and 2s must therefore be ignored when determining the number of elements and the size of the power set.

A five-element subset is obtained by selecting 5 of the 6 elements of A. The number of selections is C(6,5) = 6!/(5!1!) = 6. Equivalently, each such subset is formed by omitting exactly one element from A, and there are six possible elements to omit. Hence P(A) contains exactly six subsets with five elements.

If a finite set has n elements, then its total number of subsets, including the empty set and the set itself, is 2^n. Here 2^n = 256. Since 256 = 2^8, it follows that n = 8. The neighboring choices do not work: 2^7 = 128 and 2^9 = 512. Therefore, A has exactly eight elements.

Since A is a subset of B, every element of A is also an element of B. Therefore, an element that is not in B cannot be in A. The complement B′ contains all elements of the universal set U that are outside B, while A′ contains all elements outside A. Hence every element of B′ is also in A′, so B′ ⊆ A′. This is the reversal property of complementation.

The integers from −2 through 2 are −2, −1, 0, 1, and 2, so A has 5 elements. A set with n elements has exactly 2ⁿ subsets because each element can independently be included or excluded. Therefore, |P(A)| = 2⁵ = 32. The answer is consequently option C, not 16 or 64, which would result from counting the elements of A incorrectly.

For a subset of A, the inclusion of 1 is compulsory and the exclusion of 4 is compulsory. Thus only 2 and 3 remain undecided. Each of these two elements has two independent choices: it may be included or omitted. Therefore the number of permitted subsets is 2 × 2 = 2² = 4. They are {1}, {1,2}, {1,3}, and {1,2,3}.

At least 4 elements means that the subset may contain exactly 4 elements or exactly 5 elements. The number of four-element subsets is C(5,4) = 5, because one element is omitted. The only five-element subset is A itself, counted by C(5,5) = 1. Hence the total is 5 + 1 = 6, so option B is correct.

The elements 2 and 5 must be present in every allowed subset, so their choices are fixed. The remaining elements 1, 3, 4, and 6 are unrestricted. Each of these four elements can independently be included or excluded, giving 2⁴ possible choices. Therefore the number of subsets containing both 2 and 5 is 16, which is option B.

A subset that contains neither 1 nor 2 can use only the remaining elements 3, 4, and 5. Each of these three elements may independently be selected or not selected. Therefore the number of such subsets is 2³ = 8. This includes the empty set, single-element subsets, larger permitted subsets, and {3,4,5}; hence option B is correct.

A has three elements, so its power set contains 2³ = 8 subsets. Every subset of A belongs to P(A), but a proper subset must not be equal to A itself. Exactly one subset, namely {x,y,z}, is not proper. Removing it from the eight subsets leaves 8 − 1 = 7 proper subsets. Therefore option C is correct.

The union contains each distinct element from A and B only once. Thus A ∪ B = {1,2,3,4}, which has four elements; the repeated element 3 is counted only once. A set with four elements has 2⁴ subsets. Therefore P(A ∪ B) contains 16 elements, so option B is correct.

To form a subset of A that must contain 3, include 3 first. The remaining elements, 1 and 5, can each be either included or excluded independently. Thus there are 2 choices for 1 and 2 choices for 5, giving 2 × 2 = 2² = 4 subsets. They are {3}, {1,3}, {3,5}, and {1,3,5}. Therefore, option B is correct.

An element of P(A) must be a subset of A, so every member of that element must itself be one of a, {b}, or c. The sets {a,c}, {{b}}, and {a,{b}} satisfy this condition. However, {a,b} contains b, and b is not an element of A; only {b} is an element of A. Therefore {a,b} is not in P(A), making option D correct.

If A has n elements, then its power set P(A) has 2ⁿ elements. The given information is 2ⁿ = 512. Since 512 = 2⁹, comparing powers of 2 gives n = 9. The nearby values confirm this: 2⁸ = 256 and 2¹⁰ = 1024, so neither 8 nor 10 can be correct. Hence option C is the answer.

If B has n elements, its total number of subsets is 2ⁿ. Since 1024 = 2¹⁰, B has 10 elements. A proper subset is any subset other than the set itself. The total collection P(B) includes B once, so removing that one non-proper subset gives 1024 − 1 = 1023 proper subsets. Therefore, option A is correct.

An element of P(A) must be a subset of A. Option B, {r,t}, contains only r and t, and both are elements of A, so it is a subset and therefore belongs to P(A). Option A is an element of A rather than a set of elements; options C and D contain v, which is not in A. Hence option B is correct.

A power-set element must be a subset of A. The elements of A are 0, {1}, and 2; importantly, 1 itself is not an element of A. Options A, B, and C use only members of A and are therefore subsets. Option D contains 1 instead of the element {1}, so {0,1} is not a subset of A and is not in P(A).

The set A has two distinct elements: the empty set ∅ and the singleton set {∅}. Although one is formed from the other, they are not equal; ∅ has no elements, whereas {∅} has one element, namely ∅. Therefore |A| = 2, and the power set has 2² = 4 elements: ∅, {∅}, {{∅}}, and {∅,{∅}}. Option C is correct.

The set \(\{a\}\) has one element, so its power set has \(2^1=2\) elements: \(\varnothing\) and \(\{a\}\). Taking the power set once more gives a set with \(2^2=4\) elements. Therefore, \(|\mathcal{P}(\mathcal{P}(\{a\}))|=4\), so option B is correct. In general, if \(|A|=n\), then \(|\mathcal{P}(\mathcal{P}(A))|=2^{2^n}\).

The positive divisors of 24 are \(1,2,3,4,6,8,12,24\), so \(n(A)=8\). Alternatively, from \(24=2^3\times3^1\), the divisor-count formula gives \((3+1)(1+1)=8\). A set with 8 elements has \(2^8=256\) subsets. Therefore \(n(\mathcal{P}(A))=256\), and option C is correct.