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.
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The combinatorics of this problem look particularly simple, so let's see how much we can progress by tackling it head-on:
There are $\binom{nr}{r,r,\ldots,r} = \frac{(nr)!}{(r!)^n}$ equally likely ways to partition $nr$ objects into groups of size $r,r,\ldots,r.$
There are $n! = \binom{n}{1,1,\ldots,1}$ ways to partition the first $n$ objects, one per group.
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 $n.$
Therefore the chance must be the product of the last two divided by the first: that will answer your question.
