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Almost disjoint union (Posted on 2016-08-23) Difficulty: 3 of 5
Set S has 1600 members.
Sb is a collection of 16000 subsets of S, each having 80 members.

Prove (or disprove) that there must be at least two members of Sb containing 3 or less members in common.

No Solution Yet Submitted by Ady TZIDON    
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Solution Not so hard | Comment 1 of 2
Perhaps I am not reading the problem right, but it seems obvious that the statement is false:

Pick 4 members from S. From the remaining 1596, we have C(1596,80)=‭676286380‬ different 80-elements subsets to choose from. Just take any 16000 and we have a contradiction: They all have at least 4 members in common.

  Posted by JLo on 2020-01-24 09:13:54
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