Probability that a box has m balls I am having a hard time with a math question that sounds something like this
Suppose we have n balls that are randomly distributed into m distinct boxes. And that n = $m^2$.
What is the probability that $box_i$ has exactly m balls?
I have found solutions to similar question but I am not sure how to apply them on this question. As I understand it, the number of ways to put n balls into m boxes in m^n. Then I would need to divide it by the number of ways to put exactly m balls in a box. But I am not sure how to calculate that or whether I am on the right track or not.
 A: Assuming the uniform distribution between box and balls, the probability of each ball being assigned to box $i$ is $1/m$, let's use $x$ to indicate the event that represents a ball falling in to box $i$, dropping a ball can be expressed as
$\underbrace{\frac{(m-1)}{m}}_{\textrm{box $j \ne i$}} + \underbrace{\frac{x}{m}}_{\textrm{box } i}$
Every time a ball falls in box $i$ we have a $x$ term, probabilties of simultaneous independent events is the product of the probabilities, so if we keep multiplying this, every time a ball falls in box $i$ we have a $x$
Repeat this $m^2$ times and we have
$$\left(\frac{(m-1) + x}{m}\right)^{m^2}$$
and the probability of having $k$ balls is the coefficient of the term $x^k$, given by the binomial term
$$ \frac{(m-1)^{m^2-k} m^2!}{m^{m^2}(m^2-k)!k!} $$
Replacing $k=m$
$$ \frac{(m-1)^{m\cdot(m-1)} m^2!}{m^{m^2}(m\cdot (m-1))!m!} $$
A: There are $\binom{m^2}{m}$ ways to choose $m$ balls to be in box $i$, and then we have $m^2-m$ balls left to be distributed into $m-1$ boxes. There are $(m-1)^{m^2-m}$ ways to do this.
As you say there are $m^{m^2}$ ways to distribute these balls in general.
This gives a probability of
$$\frac{\binom{m^2}{m}(m-1)^{m^2-m}}{m^{m^2}}$$
