All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars    
perplexus dot info

Home > Just Math
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    
No Rating

Comments: ( Back to comment list | You must be logged in to post comments.)
Alternative problem? Comment 2 of 2 |
If the statement was

"Prove (or disprove) that there must be at least two members of Sb having 3 or more members in common."

it would be (a little) more difficult:

An 80-elements subset contains C(80,3) triples, so 16000 such subsets contain 16000*C(80,3) = ‭1314560000‬ triples. Since there are only C(1600,3)=‭681387200‬ different triples in total, some must show up in more than one subset.
  Posted by JLo on 2020-01-24 09:18:59
Please log in:
Login:
Password:
Remember me:
Sign up! | Forgot password


Search:
Search body:
Forums (0)
Newest Problems
Random Problem
FAQ | About This Site
Site Statistics
New Comments (24)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Chatterbox:
Copyright © 2002 - 2024 by Animus Pactum Consulting. All rights reserved. Privacy Information