Power Set and Subsets

Correct answer: B. 8

Explanation: The element 2 is required, so it is fixed as included. The element 5 is forbidden, so it is fixed as excluded. The remaining elements 1, 3, and 4 are unrestricted, and each has two choices: included or excluded. Consequently, the number of valid subsets is 2^3 = 8. Therefore, option B is correct; the restrictions leave three independent elements.

Correct answer: B. 7

Explanation: The phrase “at least 5” includes subsets having exactly 5 elements and subsets having exactly 6 elements. The number of 5-element subsets is C(6,5) = 6, while the number of 6-element subsets is C(6,6) = 1. Adding them gives 6 + 1 = 7. Hence option B is correct. The empty set and smaller subsets do not satisfy the condition.

Correct answer: B. 32

Explanation: The common element 6 must be counted only once in the union. Therefore, A ∪ B = {2, 4, 6, 8, 10}, which has 5 elements. A set with n elements has 2^n subsets in its power set, including the empty set and the set itself. Hence |P(A ∪ B)| = 2^5 = 32. Therefore, option B is correct.

Correct answer: C. 16

Explanation: The complement is taken relative to U. Removing the elements of A from U gives A' = U \ A = {1, 3, 5, 7}, so A' has 4 elements. The power set of a four-element set has 2^4 subsets, because each element can either be included or excluded. Therefore n(P(A')) = 16, making option C correct.

Correct answer: A. 6

Explanation: Every subset of A is an element of the power set P(A). To form a subset B with exactly two elements, choose any 2 elements from the 4 elements of A. The number of choices is the combination 4C2 = 4!/(2!2!) = 6. Therefore, six two-element subsets belong to P(A), so option A is correct.

Correct answer: B. 7

Explanation: For a finite set A with n elements, the power set P(A) contains 2^n elements. Here 2^n = 128. Since 128 = 2^7, comparison of the powers gives n = 7. Thus |A| = 7. Notice that 2^6 = 64 and 2^8 = 256, so options 6 and 8 cannot satisfy the given condition. Therefore, option B is correct.

Correct answer: D. a

Explanation: The power set P(A) contains every subset of A, including the empty set, all one-element subsets, all two-element subsets, and A itself. Options {a,b}, ∅, and {c} are subsets of A, so they belong to P(A). However, a without braces is an element of A, not a subset of A. Therefore, a is not an element of the power set. Braces are important: {a} is a subset, whereas a is only an element.

Correct answer: B. Composite numbers and 1

Explanation: The set A contains all prime numbers from 1 to 10: 2, 3, 5, and 7. The complement A′ is taken relative to the universal set U, so it contains the elements of U that are not in A. Removing those four primes leaves {1,4,6,8,9,10}. These are the composite numbers 4, 6, 8, 9, and 10 together with 1. Since 1 is neither prime nor composite, the correct description is composite numbers and 1.

Correct answer: B. {p, q}

Explanation: For any finite set A, the original set A is itself one of the members of its power set. In the given collection, {p,q} is the largest subset and contains both elements that occur in the singleton subsets {p} and {q}. Therefore the original set must be A = {p,q}. Notice that ∅ is a subset, not the original set here, and {{p},{q}} has different elements from A.

Correct answer: B. 5

Explanation: A subset of a four-element set can have 0, 1, 2, 3, or 4 elements. “At least 3” means exactly 3 or exactly 4 elements. The number of three-element subsets is C(4,3) = 4, because one element is omitted each time. The number of four-element subsets is C(4,4) = 1, namely A itself. Hence the required number is 4 + 1 = 5, so option B is correct.

Correct answer: A. All real numbers except −2 and 2

Explanation: Solving x² = 4 gives x = 2 or x = −2, so A = {−2, 2}. Because the universal set is the set of all real numbers, A′ consists of every real number that is not −2 and not 2. It can be written as ℝ \ {−2,2}, or equivalently as all real numbers except −2 and 2. It is not limited to integers, and it is certainly not empty because infinitely many other real numbers remain.

Correct answer: C. 4

Explanation: A member of P(A) is a subset of A. To count subsets that contain 1, include 1 in every subset and independently decide whether to include 2 and whether to include 3. Each of these two remaining elements has two choices, included or excluded. Therefore the number of valid subsets is 2 × 2 = 2² = 4. They are {1}, {1,2}, {1,3}, and {1,2,3}. Hence option C is correct.

Correct answer: A. A ⊆ B

Explanation: If A ⊆ B, every element of A is also an element of B. Consequently, every subset of A is automatically a subset of B. Since P(A) and P(B) denote the sets of all subsets of A and B respectively, this proves P(A) ⊆ P(B). The converse is also true: because A belongs to P(A), the inclusion of power sets implies A ⊆ B. Thus option A gives the exact condition, whereas disjointness or an empty union is not generally required.

Correct answer: C. 8

Explanation: Taking N as the positive natural numbers, U = {1, 2, ..., 12}, so |U| = 12. The multiples of 3 in U are A = {3, 6, 9, 12}, giving |A| = 4. The complement A' contains all members of U that are not multiples of 3. Therefore, |A'| = |U| − |A| = 12 − 4 = 8. Hence option C is correct. Listing the universal set first prevents confusion about which numbers are eligible for the complement.

Correct answer: B. 4

Explanation: We are counting subsets of A. The elements 2 and 5 are compulsory, while 1 is forbidden. The remaining elements 3 and 4 are unrestricted: each may either be included or excluded independently. Thus there are 2 choices for 3 and 2 choices for 4, giving 2 × 2 = 2^2 = 4 valid subsets. They are {2,5}, {2,3,5}, {2,4,5}, and {2,3,4,5}. Therefore, option B is correct.

Correct answer: C. 8

Explanation: A has 4 elements. The odd-cardinality subsets must contain either 1 element or 3 elements. Their number is C(4,1)+C(4,3)=4+4=8. Equivalently, for every nonempty finite set, exactly half of its 2^n subsets have odd cardinality and half have even cardinality. Since |P(A)|=16, the number of odd-sized subsets is 16/2=8.

Correct answer: A. {∅,{∅},{0},{∅,0}}

Explanation: The set A has two distinct elements: the empty set ∅ and the number 0. A set with n elements has 2^n subsets, so P(A) has 2^2=4 elements. These are the empty subset ∅, the singleton {∅}, the singleton {0}, and the whole set {∅,0}. It is important not to confuse the element ∅ with the singleton set {∅}; they are different objects.

Correct answer: B. 49

Explanation: A set with 6 elements has 2^6=64 total subsets. The subsets containing exactly 2 elements are counted by choosing 2 elements from 6: C(6,2)=6×5/2=15. The required subsets are all subsets except these 15, so their number is 64−15=49. This includes subsets of sizes 0, 1, 3, 4, 5, and 6; the empty set is also an element of the power set.

Correct answer: C. 4

Explanation: We are counting subsets of A in which 1 is definitely excluded and 4 is definitely included. Therefore only 2 and 3 remain free: each may be included or excluded independently. The number of choices is 2×2=2^2=4. The four subsets are {4}, {2,4}, {3,4}, and {2,3,4}. Thus option C is correct; treating 1 or 4 as free would overcount the possibilities.

Correct answer: B. 2

Explanation: A superset of \(\{a,b\}\) must contain both \(a\) and \(b\). The only remaining element of \(A\) is \(c\), and it may either be excluded or included. Thus the two possible supersets are \(\{a,b\}\) and \(\{a,b,c\}\). In general, if \(S\subseteq A\), the number of supersets of \(S\) contained in \(A\) is \(2^{|A|-|S|}\); here this is \(2^{3-2}=2\).

Correct answer: C. 8

Explanation: A subset must contain exactly one of the elements \(1\) and \(2\). There are two choices for this part: include \(1\) but not \(2\), or include \(2\) but not \(1\). The remaining elements \(3\) and \(4\) can independently be included or excluded, giving \(2^2=4\) choices. Therefore the total is \(2\times4=8\), so option C is correct.

Correct answer: C. 6

Explanation: A set with three elements has \(2^3=8\) subsets, so \(|\mathcal{P}(A)|=8\). The set difference removes two specific elements of the power set: the empty set \(\varnothing\) and the complete set \(A\) itself. These are distinct subsets, leaving \(8-2=6\) elements. Equivalently, the remaining subsets are the nonempty proper subsets of A. Hence option C is correct.

Correct answer: B. 4

Explanation: A subset of \(A\) is disjoint from \(\{x,z\}\) only if it contains neither \(x\) nor \(z\). Therefore, its elements can be chosen only from \(\{y,w\}\). Each of \(y\) and \(w\) may independently be included or excluded, producing \(2^2=4\) subsets: \(\varnothing,\{y\},\{w\},\{y,w\}\). Hence option B is correct; the full power set has 16 elements, but only four are disjoint from the specified set.

Correct answer: B. 4

Explanation: The governing concept is the power-set cardinality rule: a finite set with n elements has 2ⁿ subsets. First find the intersection: A ∩ B = {3, 4}, so it contains n = 2 elements. Therefore, |P(A ∩ B)| = 2² = 4. Option B is correct. The values 2 and 8 result from using the wrong exponent, while 16 would correspond to four elements.

Correct answer: B. 16

Explanation: A subset of A must include 1, so that choice is fixed. It must exclude 6, so that choice is also fixed. The remaining four elements, namely 2, 3, 4, and 5, may each be either included or excluded independently. Therefore, the number of valid subsets is 2^4 = 16. Hence, option B is correct.