# 25 people form some committees following a bunch of rules

(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 $$x\in \mathbb Z/\mathbb 5\mathbb Z$$.