Power Set and Subsets

The statement |P(A)| = 256 means that P(A) itself is a set containing 256 elements. The power set of any finite set with n elements contains 2ⁿ elements, since every element has two choices: included or excluded. Therefore |P(P(A))| = 2²⁵⁶, not 256² or 2⁸.

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.

A set with three elements has 2^3 = 8 total subsets in its power set. A proper subset cannot be equal to the original set A, so remove A itself. A non-empty subset cannot be the empty set, so remove ∅ as well. Thus, the number of non-empty proper subsets is 8 − 2 = 6. Option C is correct.

The elements of A are three distinct objects: 1, the set {1}, and the set {1, 2}. To be a subset, every element of a proposed set must be one of these three objects. In option A, both 1 and {1, 2} are elements of A, so it is a subset. In the other options, 2 or {2} is not an element of A. Therefore, A is correct.

A subset having at least six elements can have exactly six elements or exactly seven elements, because A itself has only seven elements. The number of six-element subsets is C(7,6) = 7, and the number of seven-element subsets is C(7,7) = 1. Therefore, the required total is 7 + 1 = 8, so option B is correct.

The element 0 is already included, contributing one element to every selected subset. To make the total cardinality odd, we must choose an even number of elements from the remaining four elements {1, 2, 3, 4}. The possible choices are 0, 2, or 4 elements, giving C(4,0) + C(4,2) + C(4,4) = 1 + 6 + 1 = 8. Thus, option C is correct.

The power set of A has 2^4 = 16 subsets in total. A subset is disjoint from {a, d} only when it contains neither a nor d; therefore, it can be formed using only b and c. There are 2^2 = 4 such disjoint subsets. The subsets that are not disjoint are the remaining 16 − 4 = 12 subsets. Thus, option C is correct.

A has five elements, so its power set contains 2^5=32 subsets. To count subsets containing at least one of 2 or 3, use the complement method. A subset contains neither 2 nor 3 only when it is formed from the remaining three elements {1,4,5}; this gives 2^3=8 subsets. Thus the required number is 32−8=24. Equivalently, count subsets containing 2 or 3 and subtract those containing both if using inclusion-exclusion. Therefore, option C is correct.

If the power sets P(A) and P(B) are equal, then the original sets must also be equal. One direct reason is that every element of A is a singleton or belongs to a subset of A, and the equality of the power sets makes the same subsets available for B. More simply, A itself is an element of P(A), so it is an element of P(B), and similarly B is an element of P(A); the subset relations force A=B. Therefore, option A is necessary. The other statements are not generally implied.

Any required subset must contain r and s, because it must be a superset of {r,s}. The element v is forbidden. The remaining elements t and u may each be selected or not selected independently. Thus there are 2 choices for t and 2 choices for u, giving 2×2=2^2=4 valid subsets: {r,s}, {r,s,t}, {r,s,u}, and {r,s,t,u}. Therefore, option B is correct.

The condition Y=A fixes Y completely, so there is only one possible value for the second component. The remaining condition becomes X⊆A, meaning X may be any subset of the four-element set A. A set with four elements has 2^4=16 subsets, including the empty set and A itself. Each choice of X forms exactly one ordered pair (X,A). Therefore, the number of ordered pairs is 16, so option C is correct.

First find the complement of A in U. The elements of U not in A are A'={2,4,6}, so |A'|=3. The power set of any finite set with n elements has 2^n elements, because every element has two independent choices: included or excluded. Hence |P(A')|=2^3=8. The value 16 would be the size of P(A), not P(A'), so option B is the only correct answer.

Since B has five elements, its power set contains |P(B)|=2^5=32 subsets. Since A has four elements, |P(A)|=2^4=16. Because A⊂B, every subset of A is also a subset of B, so P(A) is contained in P(B). Therefore, removing P(A) from P(B) leaves 32−16=16 elements. The strict inclusion ensures that the difference is meaningful, and option B is correct.

The set A has two distinct elements: ∅ and {∅}. The empty set ∅ and the singleton set {∅} are different objects, even though one contains the other. A set with n elements has 2^n subsets, because every element can independently be selected or not selected. Therefore |A| = 2 and |P(A)| = 2^2 = 4. Thus option B is correct.

The complement of a set is defined with respect to its universal set. Here, U contains every triangle, while E contains only the equilateral triangles. Therefore E′ = U − E consists of all triangles belonging to U that are not equilateral. It may include isosceles, scalene, and right triangles, provided they are not equilateral. Hence option A is correct.

A has 7 elements, so its power set P(A) contains 2^7 = 128 subsets in total. The number of subsets having exactly 5 elements is the binomial coefficient C(7,5) = 21, because we choose 5 elements from the 7 elements of A. The required number is the total number of subsets minus these subsets: 128 − 21 = 107. Therefore option B is correct.

There are 15 numbers from 1 to 30 divisible by 2 and 10 divisible by 3. Numbers divisible by both 2 and 3, that is, divisible by 6, are counted twice; there are 5 of them. By inclusion–exclusion, |A| = 15 + 10 − 5 = 20. Since U has 30 elements, the complement has |A′| = 30 − 20 = 10 elements. Hence option A is correct.

A subset must contain exactly one of a and b, so there are 2 choices: select a but not b, or select b but not a. The element e is required, so it has only one choice and must be included. The elements c and d are unrestricted; each may be included or excluded independently, giving 2^2 = 4 choices. Therefore the total is 2 × 4 = 8, so option B is correct.

For a subset of A to have maximum element 3, it must contain 3 and must not contain 4. The elements 1 and 2 may each be included or excluded independently. Thus there are 2 choices for 1 and 2 choices for 2, giving 2 × 2 = 2^2 = 4 valid subsets. These are {3}, {1,3}, {2,3}, and {1,2,3}. Therefore option B is correct.

The elements of A are 1, 2, and the single object {3,4}. The power set P(A) contains exactly the subsets of A. Since {3,4} is one element of A, the set containing that element alone, namely {{3,4}}, is a subset of A and therefore belongs to P(A). In contrast, 3 and 4 are not separate elements of A, so the other listed sets are not necessarily subsets. Option A is correct.