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.

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