Power Set and Subsets

Every element of C, namely 3 and 4, is present in A, and A also contains 2, so C is a proper subset of A. Every element of A is present in B, while B has the additional element 5, so A is a proper subset of B. Therefore, the complete proper-subset chain is C ⊂ A ⊂ B. The other options either reverse the inclusion, claim equality, or deny a relation that is clearly true.

Because x is an integer and 0 < x² < 10, x cannot be 0, and its absolute value must be less than √10, which is approximately 3.16. The possible integer values are therefore x = -3, -2, -1, 1, 2, and 3. Their squares are 9, 4, 1, 1, 4, and 9, all strictly between 0 and 10. Hence option A gives exactly A. Option B includes 0, option C omits ±3 and includes 0, and option D includes ±4, whose square is 16.

Any subset of A that contains B must include 1 and 3; these two elements are compulsory. The remaining elements of A are 2 and 4, and each may either be included or excluded independently. Thus there are 2 choices for 2 and 2 choices for 4, giving 2 × 2 = 2² = 4 possible subsets: {1,3}, {1,2,3}, {1,3,4}, and {1,2,3,4}. Therefore option B is correct.

Each member X of A must contain both 1 and 4 because {1,4} ⊆ X. The element 2 is forbidden, so it cannot occur in X. The only remaining element from the universal set is 3, and it can be either omitted or included. Consequently, the only possibilities are X = {1,4} and X = {1,3,4}. Thus A = {{1,4}, {1,3,4}}, making option A correct. The other choices either include the forbidden element 2 or omit a required element.

For each element of a set, a subset has two independent choices: the element is either included or excluded. Thus, a set with n elements has 2ⁿ subsets in total. For n = 5, the number is 2⁵ = 2 × 2 × 2 × 2 × 2 = 32. This count includes both the empty set and the original set itself. Therefore, option C, 32, is correct; 25 and 120 come from unrelated calculations such as 5² and 5!.

If a finite set has n elements, then its total number of subsets is 2ⁿ. The question gives 2ⁿ = 64. Since 64 = 2⁶, it follows that n = 6. Therefore, A contains 6 elements, making option B correct. For comparison, a set with 5 elements has 32 subsets, one with 7 elements has 128 subsets, and one with 8 elements has 256 subsets. These checks confirm that no other option works.

To form a subset that does not contain 2, we must exclude 2 and may choose freely from the remaining five elements: 1, 3, 4, 5, and 6. Each of these five elements has two choices, included or excluded. Hence the number of valid subsets is 2⁵ = 32. The total number 2⁶ = 64 includes subsets both with and without 2, while 48 is not a power of two and 16 corresponds to only four freely chosen elements.

A subset with exactly 3 elements is formed by choosing 3 elements from the 4 elements of A. The number of such choices is C(4,3) = 4. They are {1,2,3}, {1,2,4}, {1,3,4}, and {2,3,4}. The order of elements does not matter in a set, so arrangements are not counted separately. Therefore, option B is correct.

The power set P(A) contains every subset of A, including the empty set and A itself. If a set has n elements, its power set has 2^n elements because each original element can either be included or excluded from a subset. Here n = 3, so |P(A)| = 2³ = 8. Therefore, option C is correct; 3 counts elements of A, not its subsets.

The power set contains all subsets of A. For A = {1,2}, the subsets are the empty set ∅, the singleton sets {1} and {2}, and the set {1,2} itself. Therefore P(A) = {∅,{1},{2},{1,2}}. In a power set, each subset is an element, so writing 1 and 2 without braces is incorrect. Thus option B is correct.

The element 2 is compulsory, so it has only one choice: it must be included. The element 11 is forbidden, so it also has only one choice: it must be excluded. The remaining three elements, 3, 5, and 7, can each be included or excluded independently. Therefore the number of valid subsets is 2³ = 8, so option B is correct.

A has four distinct elements. Each element can independently be either included or excluded, so the total number of subsets is 2⁴ = 16. Exactly one of these subsets is empty, namely ∅. The phrase “at least one element” excludes that empty subset, so the number of non-empty subsets is 16 − 1 = 15. Hence option A is correct.

Because 5 must not be included, remove it from consideration. We must choose exactly two elements from the remaining four-element set {1, 2, 3, 4}. The number of choices is the combination 4C2 = 4!/(2!2!) = 6. These pairs are {1,2}, {1,3}, {1,4}, {2,3}, {2,4}, and {3,4}; therefore option B is correct.

The set A has three elements: the number 1, the number 2, and the set {1, 2}. Therefore {1, 2} ∈ A is true because that whole set appears as one element of A. Also, both 1 and 2 belong to A, so every element of {1, 2} is in A; hence {1, 2} ⊂ A is true as well. Thus option A is correct.

The power set P(A) is the set of all subsets of A, including the empty set, singletons, pairs, and A itself. The set {1, 2} contains only elements of A, so it is a subset of A and therefore an element of P(A). Although 1 is an element of A, it is not itself a subset in this context. Sets containing 4 are not subsets of A, so options C and D are wrong.

A subset of \(B\) that contains \(A\) must include 1, 2, and 3, so those three elements are fixed. Only 4 and 5 remain optional. Each optional element has two independent choices: included or excluded. Therefore, the number of such subsets is \(2^2=4\). They are \{1,2,3\}, \{1,2,3,4\}, \{1,2,3,5\}, and \{1,2,3,4,5\}. Hence option B is correct.

The element a must be included, while c must be excluded. Therefore, one position is already fixed and c cannot be selected. To make a 3-element subset, we need two more elements from the remaining available elements b, d, and e. The number of ways is C(3,2) = 3. The subsets are {a,b,d}, {a,b,e}, and {a,d,e}. Hence option B is correct.

A subset with exactly two elements is formed by choosing any 2 of the 4 distinct elements a, b, c, and d. The order of selection does not matter, so we use combinations: C(4, 2) = 4!/(2!2!) = (4 × 3)/(2 × 1) = 6. Therefore, the six subsets are {a,b}, {a,c}, {a,d}, {b,c}, {b,d}, and {c,d}.

Since x is a positive natural number and x² < 10, the possible values are x = 1, 2, and 3, because 1² = 1, 2² = 4, and 3² = 9, while 4² = 16 is not less than 10. Thus A = {1,2,3}. A proper subset must be contained in A but must not be equal to A. Only {1,3} satisfies both conditions. Option A equals A, option B contains 4, and option D contains 0, so the correct answer is C.

The empty set has zero elements, so |A| = 0. A set with n elements has exactly 2^n subsets, because each element may either be included or excluded from a subset. Thus the empty set has 2^0 = 1 subset. That single subset is the empty set itself. It is incorrect to say zero subsets, because every set contains itself as a subset; two subsets would apply to a one-element set.