Probability that out of $n$ bags, at least one contains no black ball, if $n$ of $n^2$ balls are black I am working on SL Parsonson's Pure Mathematics and I haven't been able to solve this problem:
$n^2$ balls, of which $n$ are black and the rest white, are distributed at random into $n$ bags, so that each bag contains $n$ balls. Determine the probability that at least one bag contains no black ball.
The answer given in the book is $1-\frac{(n-1)!(n^2-n)!n^{n-1}}{(n^2-1)!}$.
I thought I might start with the fact that there are $\frac{(n^2)!}{n!(n^2-n)!}$ unique arrangements of the balls, and $n-1$ partitions to be placed at intervals of n to divide them into $n$ bags, but I am stuck a little after here.
 A: I'm imagining instead of bags, we have an $n \times n$ grid of dimples where we can place balls. Then we'll consider each column of the grid to correspond to a bag.
You're correct that the number of ways to place $n$ (identical) black balls and $n^2-n$ (identical) white balls on the grid is
$$ {n^2 \choose n} = \frac{(n^2)!}{n! (n^2-n)!} $$
If no column/bag has only white balls, then there must be exactly one black ball in each column. There are $n^n$ ways to do this.
So the probability that some column/bag DOES contain only white balls is the complement of that ratio,
$$ P = 1 - \frac{n^n}{n^2 \choose n} = 1 - \frac{n!(n^2-n)! n^n}{(n^2)!} $$
The given answer has canceled a few terms, reducing (or perhaps unsimplifying) that result using $n! = n (n-1)!$, $n^n = n \cdot n^{n-1}$, and $(n^2)! = n^2(n^2-1)!$
A: It might be easier to consider all the balls distinct (say, numbered from $1$ to $n^2$).
Then the total number of arrangements is
$$ \binom{n \times n}{n}\binom{n \times (n-1)}{n} \cdots \binom{n}{n}= \frac{(n^2)!}{(n!)^n} \tag 1$$
Similarly, the total number of arrangements having a black ball in each one (complementary event) is
$$n! \frac{(n^2 -n)!}{((n-1)!)^n} \tag2$$
The ratio $(2)/(1)$is
$$  \frac{n! \, n^n (n^2-n)!}{(n^2)!} = \frac{(n-1)!n^{n-1} (n^2-n)!}{(n^2-1)!}$$
