Power Set and Subsets

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.

A subset may contain some or all elements of the original set, but every element selected must itself belong to that set. Here A contains 1, 2, and the empty set ∅ as its three elements. Therefore {1, ∅} is a subset of A. Option C, {{1}}, contains the set {1}, which is not an element of A; the element 1 alone is in A.

For a finite set with n elements, every element has two choices in forming a subset: it is either included or excluded. Thus the total number of subsets of A is 2³ = 8. A proper subset is any subset that is not equal to A itself, so we remove the one non-proper subset A. Hence the number of proper subsets is 8 − 1 = 7.

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.

To prove that C is a proper subset of A, two conditions must be checked. First, every element of C must belong to A; here both 2 and 4 are in A. Second, C must not equal A; C lacks 1 and 3, so C ≠ A. Therefore C ⊂ A. Merely having fewer elements does not guarantee inclusion.

The natural numbers satisfying 2 < x ≤ 6 are 3, 4, 5, and 6, so A = {3, 4, 5, 6}. Option B, {3, 5}, contains only elements of A and is not equal to A; therefore it is a proper subset. Option A is A itself, option C contains 2, which is excluded, and option D contains 7, which is not in A.

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.

The set A has exactly two elements: ∅ and {∅}. Its subsets are ∅, {∅}, {{∅}}, and {∅, {∅}}. Option D is {∅, {{∅}}}; it contains the element {{∅}}, but {{∅}} is not an element of A. A contains {∅}, not {{∅}}. Therefore every element of option D is not in A, so option D is not a subset. Careful counting of nested braces is essential in this question.

A has three elements: the number 1, the number 2, and the set {3, 4}. A set C is a subset of A when every element of C is also an element of A. In option B, the elements are 2 and {3, 4}; both occur directly in A, so B is a subset of A. In option A, 3 is not an element of A; only the set {3, 4} is present. Similarly, 4 and {3} are not direct elements of A in option D.

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.

The element 0 is required to be in every counted subset, so its inclusion is fixed and creates only one choice. The remaining three elements, 1, 2, and 3, may each either be included or excluded independently. Thus there are 2 choices for each of three elements, giving 2 × 2 × 2 = 2³ = 8 subsets. Equivalently, exactly half of the 2⁴ = 16 subsets of A contain any particular fixed element. Therefore option C is correct.

To form a subset that does not contain a, we may choose elements only from {b, c, d, e}. There are four remaining elements, and each element has two independent choices: it may be included or excluded. Therefore, the number of possible subsets is 2^4 = 16. This includes the empty set and every valid combination of the four allowed elements, so option B is correct.

Subset inclusion is transitive: from A ⊆ B and B ⊆ C, we obtain A ⊆ C. The additional condition A ≠ C rules out equality, so A must be a proper subset of C, written A ⊂ C. Nothing in the conditions forces B to equal either A or C, so options B and C are not always true.

A two-element subset is formed by choosing 2 elements from the 4 elements of A, and order does not matter. Hence the number is C(4,2) = 4!/(2!2!) = 6. For example, {p,q} and {q,p} are the same subset, so permutations must not be counted separately. Therefore option B is correct.