Given $N$ identical balls I have to distribute them into $n$ boxes. Any number of balls can be in one box.
What is the probability of having $l$ balls in one box?
Is this correct?
$p(l) = \frac{1}{n^{l-1}} \binom{N}{l} = \frac{1}{n^{l-1}} \frac{N!}{(N-l)!l!}$
I asked this one differently, with an explanation of how to reach to the formula: Understanding the $n!$ in the probability of $l$ events from a set of $N$ to happen at the same timestep out of $n$ total but I didn't get any feedback.
And now, based on these two related questions, I think my expression might be correct.
Probability of exactly one empty box when n balls are randomly placed in n boxes.
I think there's not a closed form for your question, since it is the number of partitions of $N-l$ balls into $n$ boxes over the partitions of $N$ balls into $n$ boxes.
Your answer is wrong for many reasons, but first of all because it is not less than 1!
For example, if $n=1$, then you have 1 box, and necessarily $p(l)=0$ for every $l\ne N$. Your formula, instead, gives $$p(l) = \binom Nl$$ that is surely distinct from 1 or zero for many values $l$.