If |A| = 6, how many elements of P(A) have cardinality 2 or cardinality 4?
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Answer and explanation
Correct answer: 30
A subset of size 2 can be selected from a six-element set in C(6,2) ways, and a subset of size 4 can be selected in C(6,4) ways. These two classes do not overlap because a subset cannot have both sizes simultaneously. Thus the required number is C(6,2) + C(6,4) = 15 + 15 = 30.
Frequently asked questions
What is the correct answer to this question?
30
Why is this the correct answer?
A subset of size 2 can be selected from a six-element set in C(6,2) ways, and a subset of size 4 can be selected in C(6,4) ways. These two classes do not overlap because a subset cannot have both sizes simultaneously. Thus the required number is C(6,2) + C(6,4) = 15 + 15 = 30.
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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