If A = {0, 1, 2, 3, 4}, how many elements of the power set P(A) contain 0 and have odd cardinality?
Answer and explanation
Correct answer: 8
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.
Frequently asked questions
What is the correct answer to this question?
8
Why is this the correct answer?
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.
Which subject and chapter does this question cover?
This is a Class 10 Mathematics question. Chapter: Sets. Topic: Power Set and Subsets.
