Difficult variation of the committee problem

Solution 1:

This is a well-known problem which I often meet in different forms. For given $m$ and $n$ let $c=c(m,n)$ be the largest possible number of committees. As far as I know, exact formulas for $c(m,n)$ are known only in particular cases, for small $m,n$ and two infinite series. There are ${m\choose 2}$ different pairs of organization members. On the other hand, each committee provides ${n\choose 2}$ such pairs, and no pair can be provided by two committees. This gives an upper bound $c\le\frac{m(m-1)}{n(n-1)}$. This upper bound is exact iff there exists a Steiner system $S(2,n,m)$. For particular values of $m,n$ their construction is based on finite projective planes. But such planes are over a finite field of a order $q$, which exists iff $q$ is a power of a prime, and there are no known other $q$ for which these exist a Steiner system $S(2, q, q^2)$ (or, equivalently, $S(2,q+1,q^2+n+1)$). Nevertheless, when $m$ is close to $n^2$ this approach provides a rather tight asymptotic lower bound for $c(m,n)$, because for sufficiently big $n$ there exists a prime number $n-n^{0.525}\le q\le n$ (see the paper “The difference between consecutive primes II” by
R. C. Baker, G. Harman, and J. Pintz). So estimating $c(m,n)$ for concrete $m$ and $n$ we usually pick a basic committee pattern provided by a finite projective plane and then try to improve the construction. The upper bound sometimes can be improved too by more subtle and complicated estimations. For instance, a current bounty question asks about minimum $m$ such that $c(m,10)\ge 40$. The current bounds are $74\le m\le 85$.

Solution 2:

For a lower bound, you can form roughly double the number of committees than the disjoint case. Divide the people into groups of size $\binom{n+1}2$, ignoring the leftovers. Identify the people of each group with the edges of a complete graph on $n+1$ vertices. For each vertex of this graph, form a committee consisting of the $n+1$ edges meeting at that vertex. There will be $n+1$ committees for each group of $\binom{n+1}2$ people, for a total of $$ (n+1)\cdot\left\lfloor \frac{m}{\binom{n+1}2}\right\rfloor\approx \frac{2m}n\;\text{ committees.} $$