Power Set and Subsets

For any finite set X, |P(X)| = 2^|X|. From |P(A)| = 32 = 2^5, we get |A| = 5. Similarly, |P(B)| = 8 = 2^3, so |B| = 3. The Cartesian product A × B therefore has |A||B| = 5 × 3 = 15 ordered pairs. Its power set consequently has 2^15 elements. Hence option A is correct.

A member of P(A) is a subset of A. If it contains no element greater than 1, its only possible elements are those not exceeding 1, namely the element 1. Thus the possible subsets are ∅ and {1}. The empty set must also be counted because it satisfies the condition, giving a total of 2 subsets.

An element of P(A) is a subset of A. Since the required subset must not contain a, a is excluded, leaving the five elements {b, c, d, e, f}. We must choose exactly three elements from these five, so the number of valid subsets is C(5,3) = 5!/(3!2!) = 10. Therefore, option A is correct.

There are 2^5 = 32 subsets in P(A). Pair every subset S with S △ {1}, where 1 is odd; toggling 1 changes the parity of the subset sum. Thus each pair contains one even-sum subset and one odd-sum subset. Consequently the 32 subsets divide equally, and the number with an even sum is 32/2 = 16.

The natural numbers in U are 1 through 50, so U has 50 elements. The multiples of 10 are {10, 20, 30, 40, 50}, giving |A| = 5. Therefore, the complement A' has 50 − 5 = 45 elements. A set with 45 elements has 2^45 subsets, so |P(A')| = 2^45. Hence, option B is correct.

A subset of {1,2,3} can use only the elements 1, 2, and 3; element 4 is not allowed. Being not a superset of {1} means that the subset must not contain 1. Therefore only 2 and 3 are available, and each may be chosen or omitted independently. The four subsets are ∅, {2}, {3}, and {2,3}.

Since A has three elements, P(A) has 2^3 = 8 elements. A singleton subset of any set contains exactly one element. Therefore, for each of the eight elements of P(A), there is one singleton subset containing it. These eight singleton subsets belong to P(P(A)), so the required number is 8.

An element of P(A) is a subset of A. Among the three specified elements 1, 2, and 3, at least two can be selected in C(3,2) + C(3,3) = 3 + 1 = 4 ways. Each of the remaining elements 4, 5, and 6 may independently be selected or omitted, giving 2^3 = 8 choices. Hence the total is 4 × 8 = 32, so option C is correct.

We need three-element subsets of A that must include 1 and must exclude 2. Since 1 is already included, one additional element must be chosen from the remaining available elements {3, 4, 5}. There are C(3,1) = 3 choices, producing {1,3,4}, {1,3,5}, and {1,4,5}. Therefore, the answer is 3, which is option B.

For a finite set A, the power set contains 2ⁿ(A) subsets. Since n(P(A)) = 64 = 2⁶, A has 6 elements. The complement A' is taken in U, so the complement rule gives n(A') = n(U) − n(A) = 10 − 6 = 4. Thus option A is correct. The value 6 is n(A), not n(A'), and 16 and 32 do not follow from the cardinality rules.

The power set P(A) contains every subset of A. The elements of A are 1, the set {2}, and 3; importantly, 2 itself is not an element of A. The set {1, {2}} uses the elements 1 and {2}, both of which belong to A, so it is a subset of A and therefore belongs to P(A). The other choices do not satisfy this condition, so option A is correct.

A set is itself an element of its power set because every set is a subset of itself. If P(A) = P(B), then A belongs to P(B), so A ⊆ B. Similarly, B belongs to P(A), so B ⊆ A. Two sets that are subsets of each other are equal. Therefore A = B is always true, making option A the only valid conclusion.

If n(A) = n, then the power set P(A) has 2^n elements. Applying the power-set rule a second time, P(P(A)) has 2^(2^n) elements. Here 2^(2^n) = 16 = 2^4, so 2^n = 4 = 2^2. Consequently n = 2, and option B is correct. The formula is applied twice, not once.

If a finite set A has n elements, then its total number of subsets is 2^n, and its number of proper subsets is 2^n − 1. Equating this to 5 gives 2^n = 6. Since 6 is not a power of 2, no non-negative integer n satisfies the equation. Therefore, no such finite set exists.

For a finite set with n elements, the power set contains 2^n subsets, including the set itself and the empty set. Hence the number of proper subsets is 2^n−1. If this number were 5, then 2^n=6, but 6 is not a power of 2. Thus no finite set can satisfy the stated condition.

The primes from 1 through 20 are 2, 3, 5, 7, 11, 13, 17, and 19, so n(A) = 8. Because the complement is relative to U, n(A') = 20 − 8 = 12. Applying the power-set rule, a 12-element set has 2¹² = 4096 subsets. Therefore option A is correct; 256, 1024, and 8192 are incorrect powers of two for this problem.

The power set lists all subsets of A: P(A) = {∅, {1}, {2}, {1, 2}}. Membership in P(A) therefore requires an object to be a subset of A. The number 1 is an element of A, but 1 alone is not a subset; {1} is the subset. Hence 1 ∉ P(A), making option A false. The empty set and A itself are always subsets of A, so options C and D are true.

A finite set with m elements has 2^m elements in its power set. Therefore, the given condition becomes 2^m=m+12. Testing the options, m=4 gives 2^4=16 and m+12=4+12=16, so the equation is satisfied. The other listed values do not satisfy it; hence m=4.

A singleton subset contains exactly one element. To form a singleton subset of A, we choose one element from the three elements of A. Thus the singleton subsets are of the form {a}, {b}, and {c} if A={a,b,c}. There are exactly 3 such subsets, and each is an element of P(A).

To form a subset with exactly two elements from a four-element set, choose 2 elements at a time without regard to order. The number is given by the combination formula C(4, 2) = 4!/(2!2!) = 6. These six pairs are all members of P(A), because every subset of A belongs to its power set.