Power Set and Subsets

The elements of P(A) are all subsets of A. We need to list subsets whose elements add to 3: {3}, {0, 3}, {1, 2}, and {0, 1, 2}. These are four different subsets. The zero element is important because adding 0 does not change a subset’s sum, so it creates an additional valid subset from both {3} and {1, 2}. No other subset has sum 3. Therefore, the correct answer is option C, 4.

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.

For A = {m, n}, its power set is P(A) = {∅, {m}, {n}, {m, n}}. A set X is a subset of P(A) only when every element of X is itself an element of P(A). In option C, the elements are m and n, but m and n are not subsets of A and therefore are not elements of P(A). The other choices contain only valid members of P(A), or are empty. Thus option C is the only set that is not a subset of P(A).

A set with n elements has exactly 2^n subsets, so |P(A)| = 2^n = 64 = 2^6. Hence n = 6. The number of 3-element subsets of a six-element set is the combination C(6, 3) = 6!/(3!3!) = (6 × 5 × 4)/(3 × 2 × 1) = 20. Therefore, option C is correct. The value 15 would count 2-element subsets, not 3-element subsets, which is a common error.

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.

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.

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.

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.

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.

P(A) is the set of all subsets of A. Therefore ∅ belongs to P(A), {1,2} belongs to P(A), and A itself belongs to P(A), because all three are subsets of A. However, 1 is an element of A, not a subset of A. The corresponding singleton set {1} is a subset and would belong to P(A), but the bare number 1 does not. Hence statement C is false.

The wording asks for subsets of \(A\) having even cardinality; these are the elements of the power set \(\mathcal{P}(A)\). Since \(|A|=5\), the number of even-cardinality subsets is \(\binom50+\binom52+\binom54=1+10+5=16\). Equivalently, for every nonempty set, even- and odd-cardinality subsets occur equally often, so each group has \(2^{5-1}=16\) subsets. Thus option B is correct.

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 \(2^{3-2}=2\).

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.

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.

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.

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.

We count subsets of A that must contain a and must not contain both b and c. The element a is fixed as included. For b and c, the allowed choices are neither, b only, or c only, giving 3 choices; the choice containing both is excluded. Elements d and e are independent, with 2 choices each. Therefore the total is 1 × 3 × 2 × 2 = 12.

Every element of A is even. Every member of P(A) is a subset of A, so it can contain no numbers other than these even elements. Since A has five elements, the number of its subsets is 2⁵ = 32, including the empty set and A itself. Therefore, 32 members of P(A) contain only even numbers.

A subset of size 2 can be selected from a six-element set in C(6,2) ways, and a subset of size 4 can be selected in C(6,4) ways. These two classes do not overlap because a subset cannot have both sizes simultaneously. Thus the required number is C(6,2) + C(6,4) = 15 + 15 = 30.

The elements of A are ∅, {a}, and b. A member of P(A) must be a set whose every element belongs to A. In option A, the elements are ∅ and {a}; both are elements of A, so {∅,{a}} is a subset of A and therefore belongs to P(A). The other choices contain elements not in A or are not subsets.