If U = {1, 2, 3, 4, 5, 6, 7} and A = {2, 4, 6}, how many subsets of P(A′) contain both 1 and 7?
Answer and explanation
Correct answer: 4
The complement is taken relative to U. Thus A′ = U − A = {1, 3, 5, 7}. We need subsets of A′ that must contain 1 and 7. Those two elements are fixed as included. The remaining elements, 3 and 5, can independently be included or excluded, giving 2 choices for each. Therefore the number of valid subsets is 2² = 4. Option B is correct; 16 counts all subsets of A′ without the required-element condition.
Frequently asked questions
What is the correct answer to this question?
4
Why is this the correct answer?
The complement is taken relative to U. Thus A′ = U − A = {1, 3, 5, 7}. We need subsets of A′ that must contain 1 and 7. Those two elements are fixed as included. The remaining elements, 3 and 5, can independently be included or excluded, giving 2 choices for each. Therefore the number of valid subsets is 2² = 4. Option B is correct; 16 counts all subsets of A′ without the required-element condition.
Which subject and chapter does this question cover?
This is a Class 10 Mathematics question. Chapter: Sets. Topic: Power Set and Subsets.
