Distributing $k$ objects in $n$ containers - Probability distribution for Combinatorial problems I have some probability notions but now I face a problem merging combinatorial and probability.
Suppose one has $k$ objects and $n$ containers with infinite capacity, $k$ and $n$ being natural integers. $k$ can be smaller, equal or greater than $n$.
The question is:
What is the probability $P(x)$ of distributing the $k$ objects within exactly $x$ containers, $x\geq0$.
Each of the $k$ objects has the same probability of falling in any of the $n$ containers, i.e., $p_0 = \frac{1}{n}$.
I do not interest myself to the number of objects in one specific container, just the distribution of the number of objects in the containers.
I understand that $\sum_{x=0}^{n} P(x) = 1$ and that $P(1)$ or $P(n)$ is ridiculously small as $k$ and $n$ increase.
I also understand how to calculate the number of different combinations using the $C_k^n = \frac{(n+k-1)!}{k!(n-1)!}$.  
By the way I am not convinced that the distribution $P(x)$ could be approached with a normal distribution as $P(x)$ will always be bounded by $x = \frac{k}{n}$ for $k>>n$.
I have been looking for a while in the different posts and two ideas came out:  


*

*The use or an hypergeometric distribution, but I don't understand why. Perhaps I misunderstood the posts suggesting it. In my idea, to use the hypergeometric distribution, one should have some sort of miss or win Bernoulli process. Source: Frequency distribution for N balls in m urns 

*The second idea I found deals with the generating function. Honestly I have never heard about that and at the point I don't know how to link it with probability distributions. What I know is that the generating function is quite helpfull to calculate moments.
If the problems solves very easily with this technique, I will naturally try to understand what generating functions are. Source: Finding generating functions for modeling the ways to distribute k different kinds of objects among n containers; 
Distributing M identical objects in N containers with capacity C and
Restricted Compositions
In all the cases, I haven't found (understood) the answer to my problem in the sources I present.
May one of you help me to solve it? Many thanks in advance for the time taken to read me and for the help you could provide.
Protra
 A: To avoid trivial cases we assume the number $k$ of objects we want to place in $x$ out of $n$ containers is greater than or equal to $x$, otherwise the probability let's denote it with $P_{n,k}(x)$ is zero.

According to OPs description we assume the $k$ objects are indistinguishable, while the $n$ containers are distinguishable.
In order to find $P_{n,k}(x)$ giving the probability that precisely $x$ containers are not empty, we calculate the number of favorable outcomes divided by the number of possible outcomes.
We start with the easy one. The number of all possible outcomes is given by the number of multisets
$$\left(\binom{n}{k}\right)=\binom{n+k-1}{k}$$
which is the number of placing $k$ objects in $n$ containers.
And now the favorable outcomes: We want to place $k$ objects in exactly $x$ containers. There are
$$\binom{n}{x}$$
possibilities to choose $x$ containers out of the total of $n$ containers.
Each of the $x$ containers has to contain at least one object. So we put into each of these containers one out of the $k$ objects, leaving $k-x$ objects. These $k-x$ objects can be placed among the $x$ containers, giving $$\left(\binom{x}{k-x}\right)=\binom{k-1}{k-x}$$ possibilities.

$$ $$

We conclude: The probability $P_{n,k}(x)$ of placing $k\geq 1$ objects in exactly $x$ out of $n$ containers, with $1\leq x \leq n$ is
  \begin{align*}
P_{n,k}(x)=\frac{\binom{n}{x}\binom{k-1}{k-x}}{\binom{n+k-1}{k}}\qquad\qquad k\geq x\tag{1}
\end{align*}
  and zero otherwise.
Since the number $x$ of containers may vary from $1$ to $n$, we obtain from (1) the  formula
  \begin{align*}
\sum_{x=1}^{n}\binom{n}{x}\binom{k-1}{k-x}=\binom{n+k-1}{k}\tag{2}
\end{align*}


[Add-on]: Let's prove formula (2) with algebraic methods to complement and verify the combinatorial proof above.
It is convenient to use the coefficient of operator $[u^x]$ to denote the coefficient of $u^x$ in a series. This way we can write e.g.
\begin{align*}
[u^x](1+u)^n=\binom{n}{x}
\end{align*}

We obtain
  \begin{align*}
\sum_{x=1}^n\binom{n}{x}\binom{k-1}{k-x}
&=\sum_{x=0}^\infty[u^x](1+u)^n[z^{k-x}](1+z)^{k-1}\tag{1}\\
&=[z^k](1+z)^{k-1}\sum_{x=0}^\infty z^x[u^x](1+u)^n\tag{2}\\
&=[z^k](1+z)^{k-1}(1+z)^n\tag{3}\\
&=[z^k](1+z)^{n+k-1}\tag{4}\\
&=\binom{n+k-1}{k}
\end{align*}
  and the claim follows.

Comment:


*

*In (1) we apply the coefficient of operator twice. We  extend the range of $x$ from $0$ to $\infty$ without changing anything since we add only zeros.

*In (2) we use the linearity of the coefficient of operator and apply the rule
\begin{align*}
[z^{k-x}]A(z)=[z^k]z^xA(z)
\end{align*}

*In (3) we use the substitution rule of the coeffcient of operator
\begin{align*}
A(z)=\sum_{x=0}^\infty a_x z^x=\sum_{x=0}^\infty z^x [u^x]A(u)
\end{align*}
and substitute $u$ in $A(u)=(1+u)^n$ with  $z$.

*In (4) we simplify the expression to $(1+z)^{n+k-1}$ and select the coefficient of $z^k$ in the last step.
