Let A = {1, 2, 3, 4, 5}. How many subsets of the power set P(A) have cardinality greater than 2?
Answer and explanation
Correct answer: 16
The set A has 5 elements, so its power set P(A) contains 2^5 = 32 subsets. We need subsets whose cardinality is greater than 2, namely those having 3, 4, or 5 elements. Their numbers are C(5,3) = 10, C(5,4) = 5, and C(5,5) = 1. Therefore, the required total is 10 + 5 + 1 = 16. Equivalently, subtract the subsets of sizes 0, 1, and 2: 32 − [C(5,0) + C(5,1) + C(5,2)] = 32 − (1 + 5 + 10) = 16.
Frequently asked questions
What is the correct answer to this question?
16
Why is this the correct answer?
The set A has 5 elements, so its power set P(A) contains 2^5 = 32 subsets. We need subsets whose cardinality is greater than 2, namely those having 3, 4, or 5 elements. Their numbers are C(5,3) = 10, C(5,4) = 5, and C(5,5) = 1. Therefore, the required total is 10 + 5 + 1 = 16. Equivalently, subtract the subsets of sizes 0, 1, and 2: 32 − [C(5,0) + C(5,1) + C(5,2)] = 32 − (1 + 5 + 10) = 16.
Which subject and chapter does this question cover?
This is a Class 10 Mathematics question. Chapter: Sets. Topic: Power Set and Subsets.
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