I have $n$ items and I would like to sort them into set of $m$ items. I would like to create these sets such that every pair of sets shares exactly $p$ elements. Additionally, all elements should be used the same number of times. Assuming this is possible with a given triplet $(n, m, p)$, what is the maximum number of sets one may construct?
A few examples: $(a, b, 0)$ yields $\frac{a}{b}$, $(q, q-1, q-2)$ yields $q$, $(n, 2, 1)$ yields $\frac{n!}{2(n-2)!}$, and $(7, 3, 1)$ yields $7$.
Thank you for any help you may provide.
The question is not so clear. Anyway, I hope this can help.
Let us fix the number of sets $N$, each of which will have $m$ items and assume that each pair of sets shares exactly $p$ items.
How many items $n$ will there be? I will say $$ n = N * m - \binom{N}{2} * p . $$ In fact, we can count $m$ items for each set and then subtract $p$ items for each pair of sets.
Furthermore, each element will be exactly in $k$ sets, where $k$ is given by $$ n * k = N * m , $$ which means $$ k = \frac{N * m}{n} = \frac{N * m}{N * m - \binom{N}{2} * p} .$$
Now, let us assume that a triplet $(n,m,p)$ is given. We can re-write the question as follows: Find $(k,N)$ such that the equations are satisfied and $N$ is maximum.