Power Set and Subsets

A one-element subset is also called a singleton set. Each of the five elements of \(A\) produces exactly one singleton subset: \(\{l\}\), \(\{m\}\), \(\{n\}\), \(\{o\}\), and \(\{p\}\). Therefore, there are five one-element subsets. Equivalently, the number of subsets containing exactly one element is \(\binom{5}{1}=5\). The value 32 is the total number of all subsets, not only singleton subsets.

The power set \(\mathcal{P}(A)\) contains exactly the subsets of \(A\). A candidate set belongs to \(\mathcal{P}(A)\) only when every one of its elements belongs to \(A\). Both 2 and 8 are in \(A\), so \(\{2,8\}\) qualifies; similarly, \(\{4,6\}\) qualifies. The empty set is a subset of every set. However, 10 is not in \(A\), so \(\{2,10\}\) is not a subset and cannot belong to the power set.

The power set is the collection of every subset of the original set. For the two-element set \(A=\{11,22\}\), the subsets are the empty set \(\emptyset\), the singleton sets \(\{11\}\) and \(\{22\}\), and the whole set \(\{11,22\}\). Thus \(\mathcal{P}(A)=\{\emptyset,\{11\},\{22\},\{11,22\}\}\), which has \(2^2=4\) elements. Option D incorrectly lists numbers as elements instead of listing the singleton subsets.

A three-element subset is formed by choosing 3 of the 4 elements \(w,x,y,z\), without regard to order. Therefore, the number is \(\binom{4}{3}=\frac{4!}{3!1!}=4\). The four subsets are \(\{w,x,y\}\), \(\{w,x,z\}\), \(\{w,y,z\}\), and \(\{x,y,z\}\). Option B, 6, counts two-element subsets using \(\binom{4}{2}\), while option D is not the required combination count.

The power set \(\mathcal{P}(A)\) is defined as the set of all subsets of \(A\). Every set is a subset of itself because each element of the set is certainly contained in that same set; thus \(A\subseteq A\). Consequently, \(A\) is one of the subsets collected in \(\mathcal{P}(A)\), so \(A\in\mathcal{P}(A)\). The other options confuse set membership with subset relation or give false conditions.

The power set \(\mathcal{P}(A)\) contains all subsets of \(A\), not all individual elements of \(A\). Since 7 belongs to \(A\), the singleton \(\{7\}\) is a subset of \(A\), and therefore \(\{7\}\in\mathcal{P}(A)\). The number 7 itself is an element of \(A\), but it is not a subset of \(A\), so it is not an element of the power set. Thus option B is correct.

The set \(A\) has five elements, and a four-element subset is formed by choosing four of those five elements. The number of choices is \(\binom{5}{4}=\frac{5!}{4!1!}=5\). Equivalently, each four-element subset is obtained by leaving out exactly one of the five elements. Thus \(\mathcal{P}(A)\) contains exactly five subsets having four elements, so option B is correct.

The power set \(\mathcal{P}(A)\) consists of every subset of \(A\). The empty set has no elements, so it cannot contain any element outside \(A\); for this reason, \(\emptyset\subseteq A\) is true for every set \(A\). Hence \(\emptyset\) is one of the subsets included in \(\mathcal{P}(A)\). It is not equal to the nonempty set \(A\), nor does it necessarily belong to \(A\) as an element.

First find the complement relative to \(U\): \(A'=U\setminus A=\{1,3,5,6\}\). Thus \(A'\) has four elements. A finite set with \(n\) elements has \(2^n\) subsets, so its power set has \(2^4=16\) elements. Therefore \(n(\mathcal{P}(A'))=16\), and option C is correct. The answer is not just the number of elements in \(A'\); it counts all its subsets.

The power set of a finite set with \(n\) elements contains \(2^n\) subsets. Here the outer elements of \(A\) are \(\{2\}\), 4, and 6, so \(n(A)=3\). The fact that \(\{2\}\) contains the number 2 does not add another outer element to \(A\). Therefore \(n(\mathcal{P}(A))=2^3=8\), and option C is correct.

The complement is obtained from the stated universal set: \(A'=U\setminus A=\{1,2,3,4,5,6\}\setminus\{3,4\}=\{1,2,5,6\}\). By definition, \(\mathcal{P}(A')\) is the collection of all subsets of this complement. Therefore the set from which those subsets are formed is \(\{1,2,5,6\}\), so option B is correct. The original set and the universal set are not the base set here.

The power set P(A) is the set of all subsets of A. Every set, including a non-empty set such as A, has exactly one empty subset: the empty set ∅. Therefore P(A) contains exactly one zero-element subset. The number 4 is the number of elements in A, while 16 is the total number of subsets because |P(A)| = 2^4 = 16; neither represents the number of empty subsets.

A power set P(A) contains every subset of A as one of its elements. The set {a, d} contains only a and d, and both of these elements belong to A = {a, b, c, d}. Therefore {a, d} ⊆ A, so it is an element of P(A). It is not the empty set, the universal set, or necessarily the complement of A.

Since A is the empty set and the complement is taken relative to U, A′ = U − A = U = {2, 4, 6}. This set has three elements. The power set of an n-element set contains 2^n subsets, because each element can either be included or excluded. Hence |P(A′)| = 2^3 = 8, making option C correct.

To form a subset of A that contains 5, keep 5 fixed and independently decide whether to include each of the remaining elements, 2 and 8. Each of these two elements has two choices: included or excluded. Therefore the number of suitable subsets is 2 × 2 = 2^2 = 4. They are {5}, {2, 5}, {5, 8}, and {2, 5, 8}.

A subset that does not contain 36 may use only the other two elements, 12 and 24. Each of these elements can be included or excluded independently, giving 2^2 = 4 possibilities. The subsets are ∅, {12}, {24}, and {12, 24}. Thus exactly four subsets of P(A) do not contain 36, so option C is correct.

A has exactly three elements. A three-element subset of A must contain all three available elements, because omitting even one element would leave a subset with only one or two elements. Thus the only three-element subset is A itself, namely {o, u, i}. Therefore P(A) contains exactly one three-element subset, and option A is correct.

The set {5} is a subset of A because its only element, 5, belongs to A. Every subset of A is an element of the power set P(A), so {5} ∈ P(A). However, 5 itself is not generally an element of P(A); the power set contains sets, not the original elements directly. Also, 7 is not in A, and P(A) is not equal to A.

Set A has two elements, so its power set contains 2^2 = 4 subsets: the empty set, the two singleton subsets, and A itself. The set P(A) therefore has 4 elements. Taking the power set once more gives |P(P(A))| = 2^4 = 16. The exponent in the second step is 4 because P(A), rather than A, is the set whose subsets are being counted.

A subset of A that contains {1, 9} must include both 1 and 9. The remaining element, 3, is optional: it may be excluded, giving {1, 9}, or included, giving {1, 3, 9}. Thus there are 2 valid subsets. Equivalently, one free element gives 2^1 = 2 choices. The question asks for subsets containing the set {1, 9}, not subsets containing only those two elements.