(Yufei Zhao) Twenty-five people form some committees with each committee has 5 members and each pair of committees have at most one common member. Determine, with justification, the maximum number of committees.

My approach: Let $m$ be the maximum number of committees. If there are more than 7 committees, then the number of duplicates is at most $\frac{m(m-1)}{2}$ but then I'm stuck. I'm guessing the answer is 10.

Am I right? Am I on the right track? Can you give me a hint if I'm not?


Each committee has $\binom{5}2=10$ pairs of people. No two committees contain the same pair of people. Since there are only $\binom{25}2=300$ pairs total, there can be at most $300/10=30$ committees.

This can be accomplished using the affine plane of order $5$. Identify the people with the set of order pairs $(x,y)$ for $x,y\in \mathbb Z/\mathbb 5\mathbb Z$. The committees are the set of "lines" over $\mathbb Z/\mathbb 5\mathbb Z$, where there $25$ lines of the form $L_{m,b}=\{(x,mx+b)\mid x\in \mathbb Z/\mathbb 5\mathbb Z\}$, where $m,b\in \mathbb Z/\mathbb 5\mathbb Z$, and $5$ vertical lines of the form $V_x=\{(x,y)\mid y\in \mathbb Z/\mathbb 5\mathbb Z\}$ for $x\in \mathbb Z/\mathbb 5\mathbb Z$.

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  • $\begingroup$ Thank you so much! That is some great insight! $\endgroup$ – Kai Jun 4 '19 at 23:26

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