I have the following problem. Let's assume I have a set of urns $\mathcal{U}=\{u_1,\ldots,u_k\}$ and a set of balls $\mathcal{B}=\{b_1,\ldots,b_n\}$. Furthermore, $m$ of these balls are special (let's suppose they are red whereas the other ones are green).
In my case, I make further restrictions. First, $k$ divides $n$, i.e., we have $\beta:=\frac{n}{k}\in \mathbb{N}$. Second, all the urns contain the same number $\beta$ of balls. Third, $k\geq m$.
We can calculate the possible number of assignments of balls to urns by $$ \phi(n,k) = \frac{n!}{(\beta !)^k} $$ (Number of all permutations divided by the product of the permutations of $k$ urns with $\beta$ balls). What I'm now interested in is the number of assignments such that at least one urn contains at least two special (or red) balls (or the number of assignments such that this does not happen, i.e., that all urns contain not more than one red ball, as I already know the total number of cases).
For the special case $\beta=2$, $m=2$, the following seems to work: $$ \frac{k\cdot\phi(n,k)}{\binom{n}{2}} $$ but I have no intuition for why this makes sense. Can anyone give me a hint?