Selfish sets

In the discrete math quiz level 2, I calculated the answer to be 386, or half of the given answer. It seems that the given solutions double-count by including a partition of the original set of 12 members into a subset of n members and its complement as a distinct partition from a subset of 12 - n members and its complement. You only need to count the different subsets with 1 to 5 members who have the property that the set and its subset are both selfish.

If I am wrong please explain my mistake.

Thanks, Fredric Kardon

#Combinatorics

Note by Fredric Kardon
5 years ago

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Comments

Can you post the link to the problem?

A Former Brilliant Member - 5 years ago

In future, if you spot any errors with a problem, you can “report” it by selecting "report problem" in the menu. This will notify the problem creator (and eventually staff) who can fix the issues.

Note that you are asked: "How many subsets have the property ...". Yes, if a subset satisfies the property then so does its complement. However, this doesn't mean that you only count them as one (pair). In your approach, after you found those with 1 to 5 members, you have to double the count.

As another example, suppose the question was "How many subsets of 5 elements have the property that both the subset and the complement are non-empty?". Yes, we only need to consider the cases when one of these subsets have 1 or 2 elements. However, we still have to double the count to properly account for the subsets which have 3 or 4 elements.

Calvin Lin Staff - 5 years ago
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