Power Set and Subsets

A subset that does not contain 1 can use only the remaining elements {2, 3, 4}. Each of these three elements has two independent choices: it may either be included or excluded. Therefore, the number of permitted subsets is 2^3 = 8. Equivalently, these are all subsets of A − {1}, including the empty set and {2, 3, 4}.

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).

A set containing n elements has 2^n subsets, because each element can either be included in or excluded from a subset. Here A has two elements, a and b. Its power set is P(A) = {∅, {a}, {b}, {a, b}}, which contains four elements. Therefore, option C is correct.

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.

The even elements in A are 2, 4, and 6. A subset formed using only even numbers may contain any selection of these three elements, including the empty subset. Each element has two independent choices—selected or not selected—so the number of subsets is 2³ = 8. Therefore, option C is correct.

The power set P(A) consists of all subsets of A. The empty set ∅ is a subset of every set, and both {∅} and {1} are also subsets of A. However, the expression {1, {∅}} contains two elements, 1 and {∅}, and both are elements of A; therefore it is a subset of A and hence belongs to P(A). It is not itself an element of A, because the elements listed in A are ∅, {∅}, and 1. Thus 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.

An element of \(\mathcal{P}(A)\) must be a subset of A. The set \(\{1,3\}\) contains only elements from A, so \(\{1,3\}\subseteq A\) and it belongs to \(\mathcal{P}(A)\). The object 1 is an element of A, not a subset by itself in this context; 4 is not in A, and \(\{\{4\}\}\) contains an element not belonging to A.

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.

The odd elements of A are 1, 3 and 5, so the relevant set has three elements. Each of these three elements has two independent choices in a subset: it may be included or excluded. Therefore the number of subsets is 2³ = 8. This count includes the empty subset, the three one-element subsets, the three two-element subsets, and the full set of odd elements. Hence option C 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.

If A has 3 elements, then its power set P(A) contains 2³ = 8 elements. A set containing 8 elements has 2⁸ = 256 subsets in total. Proper subsets are all subsets except the set itself, so we must remove one subset, namely P(A) itself. Thus the number of proper subsets is 256 − 1 = 255. Therefore option B is correct. The important point is to first find the size of P(A), and only then count its subsets.

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 \(A\subseteq X\subseteq B\).

A three-element subset is formed by choosing 3 different elements from the 5 elements of A, and the order of selection does not matter. Therefore, the required number is the combination 5C3 = 5!/(3!2!) = (5×4)/(2×1) = 10. Choosing the same three elements in another order does not create a new subset, which is why combinations rather than permutations are used.

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.