Power Set and Subsets

The element 2 is required, so it has no choice: it must be included. The element 7 is prohibited, so it also has no choice: it must be excluded. Only 3 and 5 remain free, and each can independently be included or excluded. Consequently, the number of valid subsets is 2² = 4. They are {2}, {2,3}, {2,5}, and {2,3,5}; therefore option B is correct.

An exactly three-element subset is formed by choosing 3 different elements from the 5 elements of A. Since the order of elements in a set does not matter, combinations are used rather than permutations. The required number is C(5,3) = 5!/(3!2!) = 10. Therefore, there are 10 such subsets, and option B is correct.

The set A has three elements: 1, the set {2}, and 3. The braces around 2 mean that {2} is one complete element of A, whereas the number 2 alone is not listed as an element of A. Also, a set such as {1,2} cannot be a subset because 2 is absent as an individual element. Hence only statement A is true.

The power set P(A) is the set of all subsets of A. Therefore it contains the empty set ∅, every one-element subset, every two-element subset such as {1, 2}, and A itself, {1, 2, 3}. The ordinary number 4 is not a subset of A and is not one of the listed elements of A. Hence 4 does not belong to P(A), so option D is correct.

A two-element subset is formed by choosing any 2 different elements from the 4 elements of A. The number of such choices is C(4,2) = 4!/(2!2!) = (4 × 3)/(2 × 1) = 6. These subsets are {1,2}, {1,3}, {1,4}, {2,3}, {2,4}, and {3,4}. Therefore P(A) contains 6 two-element subsets, making option B correct.

The set A has exactly two elements: ∅ and {1}. For a set to be a subset of A, every element of that set must itself be an element of A. In {∅}, the only element is ∅, and ∅ belongs to A; therefore {∅} ⊆ A. In contrast, the sole elements of {1}, {{2}}, and {0} are 1, {2}, and 0 respectively, none of which belongs to A.

Element 1 is forbidden, so it has no choice and cannot be selected. Element 6 is compulsory, so it is fixed as selected. The remaining four elements, 2, 3, 4, and 5, can each be independently included or excluded. Therefore the number of subsets is 2⁴ = 16, making option B correct.

A four-element set has 2⁴ = 16 total subsets. To count subsets containing at least one of p or q, subtract the subsets containing neither. If p and q are both absent, r and s remain free, giving 2² = 4 subsets. Hence the required number is 16 − 4 = 12. This includes subsets containing p, q, or both.

There are two mutually exclusive cases. In the first, both 1 and 2 are included; elements 3 and 4 are free, giving 2² = 4 subsets. In the second, both 1 and 2 are excluded; again 3 and 4 are free, giving 2² = 4 subsets. Adding the disjoint cases gives 4 + 4 = 8.

Taking N = {1, 2, 3, ...}, the condition x ≤ 4 gives A = {1, 2, 3, 4}. Since 4 must be included, it is fixed. Each of the remaining three elements, 1, 2, and 3, can independently be included or omitted. Therefore the number of valid subsets is 2³ = 8.

The phrase “at most two elements” includes subsets containing 0, 1, or 2 elements. From a five-element set, there is C(5,0) = 1 empty subset, C(5,1) = 5 one-element subsets, and C(5,2) = 10 two-element subsets. Therefore, the required total is 1 + 5 + 10 = 16, so option C is correct.

The governing counting principle is that subsets with at least three elements have either exactly 3 elements or exactly 4 elements. From four distinct elements, the number of 3-element subsets is C(4,3) = 4, and the number of 4-element subsets is C(4,4) = 1. Hence the required total is 4 + 1 = 5. Option B is correct. Counting only triples gives 4, while including smaller subsets would overcount categories not requested.

The condition x² = 9 means x² − 9 = 0, which factors as (x − 3)(x + 3) = 0. Thus the integer solutions are x = 3 and x = −3, so A = {-3, 3}. Since B is also defined as {-3, 3}, both sets contain exactly the same elements and therefore A = B. Hence option A is correct.

Solving x² = 16 over the integers gives x = 4 and x = −4, so A = {−4, 4}. The set C = {4} contains only 4, which belongs to A; therefore C is a subset of A. Since A also contains −4, an element not in C, the sets are not equal. Thus C is a proper subset of A.

Factor the quadratic expression: x² − 5x + 6 = (x − 2)(x − 3). Therefore x = 2 or x = 3. Both values are natural numbers, so they satisfy the restriction x ∈ N. Hence A = {2, 3}, and option A gives the set equal to A. The order of elements does not matter in a set.

The even natural-number divisors of 24 are \(A=\{2,4,6,8,12,24\}\). A set is a subset of A only when every one of its elements belongs to A. Options A and B contain only valid even divisors. Option C contains 10, and 10 does not divide 24 exactly, so C is not a subset of A. The empty set in option D is a subset of every set, including A.

The elements of A are 1, the set \(\{1,2\}\), and 3. Thus 1 is an element of A, and \(\{1,2\}\) is also an element of A, so options A and B are true. However, for \(\{1,2\}\subseteq A\), both 1 and 2 would have to be elements of A. Although 1 is in A, 2 is not a separate element of A; it occurs only inside the nested set \(\{1,2\}\). Therefore option C is false. Every set is a subset of itself, so D is true.

The set A contains three elements: 1, the nested set \(\{1,2\}\), and 3. Therefore 1 and 3 are individual elements of A, making option C true. The nested set \(\{1,2\}\) itself is also an element of A, so option B is true. But option D claims that both 1 and 2 are elements of A. The element 2 is not separately in A; it appears only inside the nested set. Hence D is the false statement.

A has four elements. To form a subset that contains the fixed element 0, we only decide independently whether each of the other three elements, 1, 2, and 3, is included or excluded. Each has two choices, so the number of subsets containing 0 is \(2\times2\times2=2^3=8\). Thus 0 appears in eight subsets in total. The number 16 is the total number of all subsets of A, including those that do not contain 0.

To count subsets that contain c, keep c fixed in every selected subset. The remaining four elements, a, b, d, and e, may each either be included or excluded independently. Thus there are 2 choices for each of four elements, giving 2^4 = 16 subsets containing c. This is also half of the 2^5 = 32 total subsets.