Power Set and Subsets

If a finite set has n elements, then each element has two choices in forming a subset: it is either included or excluded. Hence the total number of subsets is 2ⁿ. Here 2ⁿ = 16, and 16 = 2⁴. Therefore n = 4, so the set has four elements. This count includes both the empty set and the original set, as both are always subsets of a set.

Set A has four elements, so its total number of subsets is 2⁴ = 16. A proper subset is any subset that is not equal to the original set A. The original set itself is one of the 16 subsets, so it must be excluded when counting proper subsets. Thus the number of proper subsets is 16 − 1 = 15. The empty set is included among these proper subsets.

The open interval (0, 1) contains exactly the real numbers strictly greater than 0 and strictly less than 1. Both 1/4 and 3/4 satisfy these conditions, so every element of {1/4, 3/4} belongs to (0, 1). The other choices contain 0, 1, or a negative number and therefore are not subsets.

Taking N to mean the positive natural numbers, A = {1, 2, 3, 4}, so A has four elements. A set with n elements has exactly 2^n subsets, because each element has two independent choices: it may be included or excluded. Hence the number of subsets is 2^4 = 16. This count includes both the empty set and A itself.

The set A contains all real numbers greater than 1 and less than or equal to 6. For a proposed set to be a subset of A, every one of its elements must satisfy both conditions. In {2, 4, 6}, all elements are greater than 1 and 6 is allowed because the upper inequality is inclusive. The other choices contain 1, 0, or 7, which are outside A.

A set containing n distinct elements has 2ⁿ subsets, because each element has two independent choices: it may either be included in a subset or excluded from it. Here n = 2, so the number of subsets is 2² = 4. They are ∅, {1}, {2}, and {1, 2}. Both the empty set and the original set itself must be counted.

A set with two elements has 2² = 4 subsets. For {1, 2}, these are the empty set ∅, the singleton {1}, the singleton {2}, and the complete set {1, 2}. Option A lists all four and is therefore correct. Option B omits the empty and complete sets, option C lists only two subsets, and option D includes {3}, which contains an element not in the original set.

To form a subset that must contain 1, fix 1 as included. Each of the remaining three elements, 2, 3, and 4, has two independent choices: it may be included or omitted. Hence the number of subsets is 2 × 2 × 2 = 2^3 = 8. The value 16 counts every subset of A, including those without 1.

A subset that does not contain a must be formed only from the remaining elements {b, c, d}. A three-element set has 2^3 subsets because each element can independently be selected or not selected. Therefore the required number is 2^3 = 8. The number 16 represents all subsets of the original four-element set.

The power set P(A) is the set of all subsets of A. Since A = {0, 1}, its power set is P(A) = {∅, {0}, {1}, {0, 1}}. Therefore {0} is an element of P(A). The number 2 is not an element of A, so {2} and {0, 2} are not subsets of A; the standalone number 2 is not a subset either. Hence option A is correct.

The elements a and b are fixed as included in every required subset. The remaining elements c, d, and e are unrestricted: each one may either be included or omitted independently. Therefore, each of the three remaining elements gives two choices, and the total number of subsets is 2 × 2 × 2 = 2³ = 8. Hence option C is correct.

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.

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.

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.

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.