Power Set and Subsets

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.

The set has two odd elements, 1 and 3, and two even elements, 2 and 4. A subset has an even sum when it contains either zero or two odd elements. The number of choices is (choose 0 of 2 odd elements)(2^2 choices for even elements) plus (choose 2 of 2 odd elements)(2^2 choices for even elements), giving 4 + 4 = 8.

Every element of A is odd. The sum of a subset is odd exactly when the subset contains an odd number of elements. Therefore, select either one element or three elements: C(4,1) + C(4,3) = 4 + 4 = 8. The empty set and all two-element or four-element subsets have even sums, so they are not counted.

The elements of A are the numbers 1, 2, and 3, whereas the elements of B are the singleton sets {1}, {2}, and {3}. Braces are mathematically significant: 1 is not the same object as {1}. Thus A and B are not equal. Also, 1, 2, and 3 are not elements of B, while {1}, {2}, and {3} are not elements of A, so neither set is a subset of the other.

The power set P(A) consists of all subsets of A. Since A has the two elements {1} and {2}, its subsets are ∅, {{1}}, {{2}}, and {{1}, {2}}. Therefore {{1}} is an element of P(A). Notice that {1} is an element of A, not a subset of A, and 1 is neither an element of A nor a subset of A.

A two-element subset that does not contain 5 must be formed entirely from the remaining set {1, 2, 3, 4}. We therefore choose any two of these four elements. The number of choices is C(4,2) = 4×3/2 = 6. The value 10 would count all two-element subsets of A and would incorrectly include the four subsets containing 5.

The elements 2 and 5 are compulsory, so they have no choice. Element 7 is forbidden and therefore is also fixed as absent. The remaining four elements, 1, 3, 4, and 6, can each either be included or excluded independently. Hence the number of allowed subsets is 2⁴ = 16, so option C is correct.

If a finite set has n elements, its total number of subsets is 2ⁿ. A proper subset includes every subset except the set itself, so the number of proper subsets is 2ⁿ − 1. Given 2ⁿ − 1 = 31, we obtain 2ⁿ = 32 = 2⁵. Therefore n = 5, making option B correct. The empty set is included among the proper subsets.

Set B has exactly one element, and that element is the set \(\{1,2\}\) itself. Since A is the set \(\{1,2\}\), A is therefore an element of B, so \(A\in B\). This does not mean A equals B: A has two numerical elements, whereas B has one set-valued element. Also, B is not a subset of A because its element \(\{1,2\}\) is not the number 1 or 2, and 1 is not directly an element of B.

Because both p and r belong to A, every element of B belongs to A; hence \(B\subseteq A\). The subsets of B are \(\varnothing\), \(\{p\}\), \(\{r\}\), and \(\{p,r\}\). Each of these contains only elements from A, so each is also a subset of A. This is the transitive property of inclusion: if \(C\subseteq B\) and \(B\subseteq A\), then \(C\subseteq A\). Therefore all subsets of B satisfy the requirement.

A set C is a subset of A when every element of C is also an element of A. The empty set is a subset of every set, so option A qualifies. Both 0 and 2 are in A, making option B a subset, and 1 is in A, making option C a subset. In option D, the only element is the empty set \(\varnothing\), not the number 0. Since \(\varnothing\notin A\), the set \(\{\varnothing\}\) is not a subset of A.

If a finite set A has n elements, then each element has two choices when forming a subset: it is either included or excluded. Therefore the power set has \(|P(A)|=2^n\) elements. Here \(2^n=16=2^4\), so n=4. Thus A contains 4 elements. The number 16 describes the total number of subsets, not the number of original elements. For comparison, sets with 2 and 3 elements have power sets of sizes 4 and 8 respectively.

The elements of \(P(A)\) are exactly the subsets of A. Option B, \(\{2,3\}\), is a subset because both 2 and 3 belong to A; therefore it is an element of \(P(A)\). Option A is only the number 2, an element of A but not a subset of A. Option C is not a subset because 4 is absent from A. Option D contains the set \(\{1\}\) as its element, rather than the number 1, so it is not a subset of A. The distinction between membership and subset notation is essential.

If \(A\subseteq B\), every element of A belongs to B. If at the same time \(B\subseteq A\), every element of B belongs to A. Thus neither set contains an element absent from the other, so they have exactly the same elements and must be equal: \(A=B\). This is the standard two-inclusion method for proving equality of sets. The conclusion says nothing about whether the sets are finite or infinite. Their intersection is actually the common set itself, not the empty set.

A subset must contain only elements that are themselves elements of A. The elements of A are the number 1, the set {2}, and the number 3. Option B contains {2} and 3, so every element of that option belongs to A; hence it is a subset. In options A and D, the number 2 appears, but 2 is not an element of A. In option C, the element {1, 2} is not in A. The distinction between 2 and {2} is essential.

An exactly two-element subset is formed by choosing two different elements from the five elements of A. Since the order of selection does not matter, combinations are used: C(5, 2) = 5!/(2!3!) = (5 × 4)/2 = 10. Thus there are 10 such subsets. The value 20 would count ordered selections using permutations, while 5 and 25 do not represent the number of two-element subsets. Every pair is counted once, regardless of its order.

A set with four elements has 2^4 = 16 total subsets. The subsets having fewer than two elements are the empty set, which has zero elements, and the four singleton subsets {a}, {b}, {c}, and {d}; therefore there are 1 + 4 = 5 excluded subsets. The number with at least two elements is 16 - 5 = 11. Equivalently, one may add C(4,2) + C(4,3) + C(4,4) = 6 + 4 + 1 = 11.

When A is a subset of B, every element of A is already contained in B. If B had even one additional element, its cardinality would be greater than that of A. Since both finite sets have cardinality 6, B cannot contain any extra element. Therefore the two sets contain exactly the same elements, so A = B.

The condition x² = 1 can be factored as (x − 1)(x + 1) = 0. Therefore x = 1 or x = −1. Both values are integers, so the set defined by the rule is A = {−1, 1}. This is exactly the set given as B. Hence A = B. The other options omit one valid solution or incorrectly claim that there are no solutions.

The positive divisors of 36 are 1, 2, 3, 4, 6, 9, 12, 18, and 36. Among these, the perfect squares are 1 = 1², 4 = 2², 9 = 3², and 36 = 6². Therefore A = {1, 4, 9, 36}. Option B omits two valid divisors, option C incorrectly includes 6, and option D contains no perfect-square elements.

The positive multiples of 10 that are less than 50 are 10, 20, 30, and 40. Therefore, \(A=\{10,20,30,40\}\) has four elements. A set with \(n\) elements has \(2^n\) subsets in total. Exactly one of these is the set itself, so the number of proper subsets is \(2^4-1=16-1=15\). Thus option B is correct; 16 would count all subsets, including \(A\) itself.

A three-element subset is formed by choosing 3 of the 4 elements of \(A\). The number of such choices is \(\binom{4}{3}=\frac{4!}{3!1!}=4\). Each resulting set has only three elements, whereas \(A\) has four, so none of them equals \(A\); consequently, all four are proper subsets. They are \{1,2,3\}, \{1,2,4\}, \{1,3,4\}, and \{2,3,4\}.

The power set \(P(A)\) is the set of all subsets of \(A\). Since both 1 and 3 belong to \(A\), the set \(\{1,3\}\) is a subset of \(A\). Every subset of \(A\) is an element of \(P(A)\), so \(\{1,3\}\in P(A)\). Notice the distinction: 1 and 3 are elements of \(A\), while the set \(\{1,3\}\) is an element of the power set. Therefore option A is correct.

A singleton subset contains exactly one element. The singleton subsets of {a, b, c} are {a}, {b}, and {c}. Therefore, A = {{a}, {b}, {c}}. This is exactly the given set B, so A and B have the same elements and A = B. Notice that a, b, and c themselves are not the elements of A; the singleton sets {a}, {b}, and {c} are its elements. Thus neither set is a proper subset of the other.

If A has n elements, then its power set has 2ⁿ elements. Given |P(A)| = 64, we have 2ⁿ = 64 = 2⁶, so n = 6. A set with six elements has 2⁶ = 64 total subsets. Exactly one of these subsets is A itself, and a proper subset must be different from A. Therefore the number of proper subsets is 64 − 1 = 63. The empty set is included among the proper subsets.