Power Set and Subsets

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.

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.

The power set P(A) consists of all subsets of A. For A = {1, 2}, we have P(A) = {∅, {1}, {2}, {1, 2}}. Each member of P(A) is itself a set. The object 1 is an element of A, but the singleton set {1} is an element of P(A); these are different objects. Therefore 1 is not an element of P(A).

The odd elements of A are 1 and 3. Exactly one of these two must be selected, which can be done in C(2,1) = 2 ways. The even elements 2 and 4 have no restriction, so each may be included or excluded independently, giving 2² = 4 choices. Therefore the total number is C(2,1) × 2² = 2 × 4 = 8. Hence option D is correct.

For a set with n elements, each element has two independent choices when forming a subset: it is either included or excluded. Therefore the total number of subsets is 2^n. With n = 3, the number is 2^3 = 8. This count includes the empty set and the original set itself. The value 7 would omit one of these, while 6 counts only non-empty proper subsets. Hence option C is correct.

The element 2 must be included in every required subset, so it is not a choice. The remaining elements 1, 3, and 4 can each independently be included or excluded. Thus there are 2 choices for each of three elements, giving 2³ = 8 subsets. Equivalently, a four-element set has 2⁴ = 16 total subsets, and exactly half contain any fixed element, so 16 ÷ 2 = 8.

To form a subset that does not contain p, p is excluded permanently. The remaining four elements q, r, s, and t are each free to be included or excluded independently. Therefore the number of valid subsets is 2⁴ = 16. Another way to see this is that all subsets of the four-element set {q, r, s, t} are exactly the subsets of A that omit p. The empty subset is included in this count.

Equal sets have exactly the same elements and therefore exactly the same subsets. For any set X, \(X\in\mathcal{P}(A)\) means \(X\subseteq A\). Since A and B are equal, this is equivalent to \(X\subseteq B\), which means \(X\in\mathcal{P}(B)\). Thus the two power sets have identical elements and are equal. They are not generally empty.

The elements 2 and 7 are required in every counted subset, so they are fixed and do not create choices. The remaining elements, 3 and 5, may independently be included or omitted. Thus there are 2 choices for 3 and 2 choices for 5, giving 2 × 2 = 4 subsets. They are {2,7}, {2,3,7}, {2,5,7}, and {2,3,5,7}. Therefore option B is correct.

A set with n elements has 2ⁿ total subsets because every element can either be selected or not selected. Here n = 4, so A has 2⁴ = 16 subsets. Exactly one of these is the empty set, which has no elements. Therefore, the number of subsets containing at least one element is 16 − 1 = 15. Hence option B is correct; option C includes the empty set and is therefore too large.

The power set P(A) is the set of all subsets of A. The set {∅, 1} contains two elements, ∅ and 1, and both of them are elements of A; therefore, {∅, 1} is a subset of A and belongs to P(A). However, {∅, 1} itself is not listed among the three elements of A. In contrast, 1, ∅, and {∅} are all elements of A. Hence, option B is the only correct answer.

Every valid set \(X\) must contain the elements of \(A\), namely 1 and 2, and it cannot contain anything outside \(B\). The optional elements are therefore \(B\setminus A=\{3,4,5,6\}\). Each of these four elements can independently be either included in or excluded from \(X\). Consequently, the number of possible sets is \(2^4=16\), so option C is correct. This is the general formula \(2^{|B|-|A|}\) when \(A\subseteq X\subseteq B\).

The conditions fix the status of 1 and 5: 1 must be included and 5 must be excluded. The remaining elements 2, 3, and 4 are unrestricted, and each has two independent choices—either it is included or it is not. Hence the number of valid subsets is 2 × 2 × 2 = 2^3 = 8. No additional factor is needed because the required and forbidden elements are already fixed.

A set with four elements has 2^4 = 16 total subsets. To count subsets containing at least one of a or b, use the complement: subsets containing neither a nor b can use only c and d, so there are 2^2 = 4 such subsets. Therefore, the required number is 16 − 4 = 12. This method avoids double-counting subsets that contain both a and b.

For a finite set A, the number of elements in its power set is given by |P(A)| = 2^|A|. Since |P(A)| = 32 and 32 = 2^5, we have 2^|A| = 2^5. Equality of the powers gives |A| = 5. The other options would produce 16, 64, or an enormously larger power-set size, so they cannot be correct.

Every member of \(\mathcal{P}(A)\) is a subset of \(A\). Since \(A\subseteq B\), any subset of \(A\) is automatically also a subset of \(B\). Therefore, every element of \(\mathcal{P}(A)\) belongs to \(\mathcal{P}(B)\), giving \(\mathcal{P}(A)\subseteq\mathcal{P}(B)\). Equality occurs only when \(A=B\); in general, \(B\) may have additional elements and therefore additional subsets.

For a finite set with \(n\) elements, the power set contains \(2^n\) subsets. Therefore, \(|\mathcal{P}(B)|=2^3=8\) and \(|\mathcal{P}(A)|=2^2=4\). Because \(A\subset B\), every subset of \(A\) is a subset of \(B\), so \(\mathcal{P}(A)\subseteq\mathcal{P}(B)\). The difference therefore has \(8-4=4\) elements. Thus option B is correct.

The odd elements of \(A\) are \(1,3,5\), so there are three odd elements. To obtain a subset with exactly two odd elements, choose two of these three odd elements: \(\binom{3}{2}=3\) choices. Each of the three even elements, \(2,4,6\), may be independently included or excluded, giving \(2^3=8\) choices. By the multiplication principle, the total is \(\binom{3}{2}2^3=3\times8=24\).

A subset with at least 4 elements can have either exactly 4 elements or exactly 5 elements. The number of 4-element subsets is C(5,4) = 5, and the number of 5-element subsets is C(5,5) = 1. Therefore, the required number is 5 + 1 = 6. Hence option B is correct. The answer 5 counts only the four-element subsets and misses the whole set.

Any subset that does not contain a can use only the remaining four elements b, c, d, and e. Each of these four elements may either be included or excluded, so the number of such subsets is 2^4 = 16. None of them can equal A because A contains a, so every one of these 16 subsets is proper. Therefore option C is correct.

For integer x, the inequality x² ≤ 4 gives −2 ≤ x ≤ 2. Hence A = {−2, −1, 0, 1, 2}, which has five elements. A set with n elements has 2^n total subsets, so A has 2^5 = 32 subsets. A proper subset is any subset other than the set itself, so the number of proper subsets is 32 − 1 = 31. Therefore option B is correct.

The power set P(A) contains every subset of A as an element. Since A = {1, 2, 3}, its elements include ∅, {1}, {2}, {3}, {1,2}, {1,3}, {2,3}, and A itself. In option A, ∅, {1}, and {2,3} are all subsets of A, so the whole set in option A is a subset of P(A). The other options contain 1 or 4 directly, whereas those are not necessarily elements of P(A) in the required form; 4 is not even an element of A.

First find the complement of \(A\) in the universal set: \(A'=U-A=\{3,4,5,6\}\). Thus, \(A'\) has 4 elements. For any finite set with \(n\) elements, the power set has \(2^n\) elements, because each element can independently be selected or not selected. Therefore, \(|\mathcal{P}(A')|=2^4=16\).