# Conditional Probability of balls in cells

Suppose that we have nr balls where $$n \geq 2,r \geq 2$$. The balls are numbered $$1, 2, \ldots, nr$$. The balls are placed into n cells with no restriction on the number of balls allowed per cell. Assume each ball is placed into a cell independently of other balls. Given that each of the n cells has exactly r balls. Find the conditional probability that the balls numbered $$1,2, \ldots, n$$ are in different cells. Let D denote the event that the balls numbered $$1,2, \ldots, n$$ are in different cells and let $$E$$ denote the event that each cell has exactly $$r$$ balls. So we are looking for: $$P(D|E)= \frac{P(D \cap E)}{P(E)}$$ I have found that $$P(E)=\frac{(nr)!}{(r!)^nn^{nr}}$$ I am having trouble finding $$P(D \cap E)$$, does anybody have any ideas on who to find this.

1. There are $$\binom{nr}{r,r,\ldots,r} = \frac{(nr)!}{(r!)^n}$$ equally likely ways to partition $$nr$$ objects into groups of size $$r,r,\ldots,r.$$
2. There are $$n! = \binom{n}{1,1,\ldots,1}$$ ways to partition the first $$n$$ objects, one per group.
3. There are $$\binom{n(r-1)}{r-1,r-1,\ldots,r-1} = \frac{(n(r-1))!}{((r-1)!)^n}$$ ways to partition the remaining $$n(r-1)$$ objects into groups of size $$r-1$$ independently of the partitioning of the first $$n.$$