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We have $40$ meetings, each one containing $10$ people. There aren't any two people who have been in more than $2$ same meetings. If $n$ is the number of people, prove that $n$ is at least $82$.

So far double counting. I've proved that it is bigger than $67$. I also tried graph and inequalities about vertices and edges, but I couldn't prove $n>82$.

I took every $10$ people together and by double counting , since each $10$ people can cover at most $58$ meeting, I concluded that $n>67$. I tired taking $11$ people but it didn't work and got complicated.

I also tried to calculate the number of $2$-paths of a Bipartite graph Corresponded to the problem which number of vertices of one part is $n$ and the other part $40$. And using $$\sum_i^{n+40} \binom{d_i}{2} \geq \frac{2m^2}{n+40}-m$$ that we get $m=400$. But it didn't work too.

And since I'd proved $n>67$ I tried to set such a Corresponding Bipartite graph with $n=68$, but I have no idea how to make such a graph.

According to the problem we get the number of $4$-cycles of the graph is at most $\binom{n}{2}$, but I don't know what to do with it.

Totally I believe the main or only solutions is by counting $2$-paths of the corresponding graph!

Parzival
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  • What have you tried? – Jens Mar 12 '19 at 18:38
  • When providing new information, please click on the tiny edit and use MathJax and improve the body of the post instead of commenting. – Lee David Chung Lin Mar 12 '19 at 23:55
  • "I don't know why my question is on hold." -- I voted to put your question on hold because it was a PSQ (problem statement question) at the time I saw it. You literally only posted the problem itself - you posted nothing about the context in which it came up, about your attempts, or about where specifically you had trouble. (Others may have voted for other reasons but those are the most important to me.) [cont.] – PrincessEev Mar 13 '19 at 06:46
  • MSE discourages the posting of such, because MSE is not a "do my homework for me" type of site; if you expect to get help on homework, at least here, you should make a genuine effort to learn and put in effort to it just as the people here do when they spend time trying to help you. ... With respect to your edits, I see that they have drawn at least three reopening votes. Personally, I don't feel inclined to put my own vote into the pool until you elaborate on where your results come from and the attempts you made. But that's me. – PrincessEev Mar 13 '19 at 06:48
  • I know what you mean. I didn't do those things because i didn't know i had to do. – Parzival Mar 14 '19 at 13:55
  • So for any pair of people they can be together on at most 2 meetings? – nonuser Mar 21 '19 at 05:11
  • I'm sorry but I think that 82 is way optimistic. I don't know if even that 68 is true. – nonuser Mar 21 '19 at 05:31
  • First one yes,and about 68 i'm sure. – Parzival Mar 21 '19 at 09:36
  • This turns out to be a repeat of: https://math.stackexchange.com/questions/3023032/finding-the-minimal-number-of-members/3037615#3037615 -- where the (short & sweet) answer by Song shows $n \ge 82$ and my (much more tedious) answer constructs an $n = 82$ solution by using a specific Steiner system. I remember having a lot of fun solving this problem. Enjoy! :) – antkam Mar 22 '19 at 19:18

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