A little quirk on the good old "how many ways are there to place balls into boxes" Given $n$ (unlabeled) balls, and $m$ (labeled) containers of equal size, I want to calculate what is the most probable distribution of balls, from the point of view of have many containers will remain empty. Balls are assigned randomly , with equal probabilites for each container.
So, I believe the number of configurations in which $k$ containers out of the available $m$ contain at least one ball equals
$$ C(k) = { {n -1} \choose {k -1} }{ {m} \choose {k} } $$ 
where the first term gives the number of combinations to put $n$ balls in $k$ containers with each container containing at least one ball (Put $N$ identical balls into $m$ different buckets, each bucket has at least one ball, how many ways?), and the second reflects the freedom in choosing the $k$ containers to place balls in.
It is a matter now to maximise the function $C(k)$. 
I checked that when Stirling's approximation applies, for large arguments of the factorial, the maximum tends to $n$, so filling all the boxes. In other words, with many particles, it is unlikely any box will remain empty.
But plotting $C(k)$ for smaller $n$ and $m$, one notices the maximum occurs for $k<m$, so some containers are indeed empty.
Now, the question.
I am interested in estimating when the transition occurs, i.e. for which values of $m, n$ the maximum of $C(k)$ occurs for $k < m$. I am stuck a bit as I cannot manipulate the binomial coefficients and the factorials for small values of their argument.
My attempt followed this route.  Let us assume that with given $n$ and $m$, one computes the most likely outcome is such that $k$ containers only are filled. But then, one could divide the containers in two sets, filled and empty by the process described above, each considered as a container in itself. One can view the process as a Bernoulli process, in which balls are assigned to a larger Container, union of  the selected $k$ "filled" containers) with probability proportional to $k$ (and this is a Bernoulli "success"), and the union of the $m-k$ empty containers, with probability proportional to $m-k$. The analogue of getting $k$ empty containers in the above described process is, for the Bernoulli process, to get all successes over the $n$ attempts.  For the event "all successes" to be the expected value, the following conditions has to hold
$$ n p \geq n -1/2 $$
where $p$, the chance of success on the single  draw, equals 
$$ p = \frac {m-k}{m}$$
I end up then with the condition
$$ n (\frac {m-k}{m} ) \geq n -1/2 $$
At least, I believe this confirms that when $n \to \infty$, $k \to 0$.
Any hint on how to substantiate the condition found with a direct calculation on $C(k)$ would be greatly appreciated, thanks.

This isn't a complete answer to the OP's question, but it corrects the expression given for the required probability.
Let $P(n,m,k)$ be the probability that after $n$ balls are randomly assigned to $m$ containers, exactly $k$ of them have at least one ball in them.
Using the inclusion-exclusion formula, we find
$$P(n,m,m) = 1 - m(1-1/m)^n + \binom{m}{2}(1-2/m)^n - \dots = \sum_{i=0}^{m-1} (-1)^i \binom{m}{i}(1 - i/m)^n.$$
Now, summing over all possible $k$-element subsets of the $m$ containers, we have 
$$P(n,m,k) = \binom{m}{k}(k/m)^nP(n,k,k) = \binom{m}{k} \sum_{i=0}^{k-1} (-1)^i \binom{k}{i}\left(\frac{k-i}{m}\right)^n =  \frac{m!}{m^n(m-k)!}S(n,k),$$
where $S(n,k)$ denotes a Stirling number of the second kind.
