M indistinguishable balls on N indistinguishable boxes Through some research I found that the answer is $\Omega = \binom{M+N-1}{N-1}$
But why? I found an explanation which explained it like this:
Let the balls be  $\circ$. To find out how the balls are distributed in the boxes we use $N-1$ "|". That way we have $M+N-1$ symbols If the boxes and balls were distinguishable we would have $(M+N-1)!$ combinations. Since they are distinguishable we have to divide this by $(N-1)!\cdot M!$ . Example: $M=3$ and $N=2$
|$\circ$$\circ$$\circ$
$\circ$|$\circ$$\circ$
$\circ$$\circ$|$\circ$
$\circ$$\circ$$\circ$|
Could someone maybe explain this in more detail? I don't really understand why this works. Why can we use $N-1$ symbols to find out the number of combinations.  I read a lot on this but I couldn't find an explanation which made things clear for me yet.
I am open for a completely different approach too. 
 A: Say I give you $M$ red balls numbered $1, 2, \ldots, M$ and $N-1$ green balls numbered $1, 2, \ldots, N-1$.  How many ways are there to arrange these in a line?  Clearly, each ball is distinuigshable by their combined attributes of color and number, so there are simply $(M+N-1)!$ ways to arrange these balls.
Now, suppose the balls retained their color but they are no longer numbered.  Then how many distinct arrangements are possible?  We reason that for each such arrangement, there are $M!$ admissible numberings of the red balls, and $(N-1)!$ admissible numberings of the green balls.  Since the numberings of the balls in one color group can be performed independently of the numbering of the other color group, it follows that each desired un-numbered arrangement corresponds to $M! (N-1)!$ numbered arrangements that we counted in the first scenario above.  Thus the desired number of arrangements of the colored balls with no numbers is $$\binom{M+N-1}{M} = \binom{M+N-1}{N-1}.$$
Now, why are there $N-1$ green balls in the first place?  What is the correspondence of this example to the case where balls are placed in boxes?  The answer is that each distribution of balls in boxes in the original question can be identified with a partition of the balls when they are arranged in a line.  For example, if I have 3 boxes and 7 identical balls, I could place the 7 balls in a line like so:  $$*******$$  Then to represent how these balls are placed in the boxes, I simply partition these balls; e.g., $$*** \mid * \,* \mid ***$$ means I have put three balls in the first box, two in the second, and three in the third.  I don't need a third division mark because once I've used two, the number of groups of balls is now 3.  That's why there are $N-1$ marks in the general case of $N$ boxes.
Now whether this enumeration actually corresponds to the title of your question is a different matter, since you are saying the boxes are also indistinguishable.  In such a case, you could not tell the difference between $$*** \mid * \,* \mid ***$$ and $$*** \mid *** \mid * \,*$$ because both of them have two boxes with three balls, and one box with two.  Indistinguishable boxes, then, implies that the order of the boxes themselves are not taken into account.  If this is what you want, then the binomial coefficient $$\binom{M+N-1}{M}$$ is not the desired number of arrangements.  The correct number is the number of unordered partitions of the number $M$ into exactly $N$ nonnegative integers.
A: The number of ways of distributing $M$ identical objects into $N$ identical boxes is equal to the number of ways of writing the integer $M$ as a sum of at most $N$ positive integers, ordering not taken into account and with repetition allowed. Its ordinary generating function is
$$\frac{1}{(1-x)(1-x^2) \cdots (1-x^N)}$$
This result can be found in most texts which discuss generating functions in combinatorics. Note that the answer is not
$$\binom{N+M-1}{M} = \binom{N+M-1}{N-1}.$$
The latter is actually the number of ways of distributing $M$ identical objects into $N$ distinct boxes.
