If |A| = 5, how many members of 𝒫(A) contain a fixed element a ∈ A and have odd cardinality?
Answer and explanation
Correct answer: 8
The fixed element a is already included, contributing one element. To make the complete subset have odd cardinality, we must select an even number of elements from the remaining four elements. Among four elements, the number of subsets with even size is 2^(4−1) = 8. Equivalently, half of all 2^4 choices have even size. Therefore, eight members of 𝒫(A) satisfy both conditions, so option B is correct.
Frequently asked questions
What is the correct answer to this question?
8
Why is this the correct answer?
The fixed element a is already included, contributing one element. To make the complete subset have odd cardinality, we must select an even number of elements from the remaining four elements. Among four elements, the number of subsets with even size is 2^(4−1) = 8. Equivalently, half of all 2^4 choices have even size. Therefore, eight members of 𝒫(A) satisfy both conditions, so option B is correct.
Which subject and chapter does this question cover?
This is a Class 10 Mathematics question. Chapter: Sets. Topic: Power Set and Subsets.
