Consider the problem : bakery make a dough for $m$ cakes after this chef-cook add $n$ cherries to dough. After cooking we have $m$ cakes. Suppose we randomly select one cake, what is the probability that this cake has $k$ cherries?
My attempt (edited) : I understand that my previous approach is bad.
Consider numerical example : $m=3,n=2,k=1$. So we have only $2$(!) possible variants of cherries positions $(2,0,0)$ and (1,1,0) (because we don't need to enumerate the cakes, so the cakes are non-ordered , i.e. $(0,2,0)$ = $(2,0,0)$), . Now we select one cake, there are only $2$ cakes with $1$ cherrie inside. But all number of cakes is : $6$, so the probability is : $\frac{2}{6}= \frac{1}{3}$.
If I understand what you are saying, then I think you have a problem. I'm going to restate the question here so you can see if I understand you.
We have $m$ cakes, and there are $n$ cherries randomly distributed throughout the cakes, in the sense that each cherry is equally likely to be in any one of the cakes. A cake is chosen uniformly at random. What is the probability that it contains exactly $k$ cherries?
Now you say, let $a_i$ be the number of cherries in cake $i.$ There are $\binom{n+m-1}{n-1}$ such sequences, and each is equally likely. Suppose our chosen cake has exactly $k$ cherries. Then the remaining $m-1$ cakes have $n-k$ cherries, and there are $\binom{n+m-k-2}{n-k-1}$ ways for that to happen, so the answer is $$\frac{\binom{n+m-k-1}{n-k-2}}{\binom{n+m-1}{n-1}}$$
The problem is that for this to work, each distribution of cherries has to be equally likely. That is, if we have three cakes, and 20 cherries, it's just as likely that all the cherries are in the first cake as it is that 7 are in the first cake, 6 in the second, and 7 in the third, and that's not so.
I think you want to reason like this. Each cherry has probability $1/m$ of being in the chosen cake, and probability $(m-1)/m$ of being in some other cake. So, we're looking for the probability of exactly $k$ successes in $n$ Bernoulli trials, with probability of success $1/m$.