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I was asked to obtain 6 balls in 4 boxes case with each ball and box to be the same. Then I obtain there will be 8 cases, say (6,0,0,0), (5,1,0,0), (4,2,0,0),(4,1,1,0),(3,3,0,0),(3,2,1,0),(3,1,1,1),(2,2,1,1) (order is not a matter).

So I am thinking, can one generalize the case for N indistinguishable balls in M indistinguishable boxes. How many cases are? It can be interpreted as how many different equations can I have with$$x_1+x_2+x_3+...+x_M=N$$ and for each $x_i\ge0, x_i\ge x_j,\forall i\ge j$.

I still have no idea about that.

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    https://en.wikipedia.org/wiki/Partition_(number_theory) – 2017-01-27
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    Partitions of $N$ into at most $M$ parts. – 2017-01-27
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    "*It can be interpreted as how many different* **solutions** *with $x_1+x_2+\dots+x_M = N$*" no, this is a different question. You would need to enforce the extra condition that $i\geq j\implies x_i\geq x_j$, else the order of the numbers would matter too. That would have been for if the boxes were distinct, not identical. – 2017-01-27
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    @JMoravitz Yes, you are right. But you know what I mean : ) – 2017-01-27
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    Possible duplicate of [Number of ways of putting n indistinguishable balls into k indistinguishable groups.](http://math.stackexchange.com/questions/1171473/number-of-ways-of-putting-n-indistinguishable-balls-into-k-indistinguishable-gro) – 2017-01-27

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If you are looking for a formula (though it is difficult to calculate), here it is.

Required number of ways is

$$\sum_{i=0}^4 P(6,i)$$

Where $P(k,n)$ = number of integer partitions $k$ into $n$

and $$P(k,n)\approx\dfrac{1}{4n\sqrt{3}}e^\left(\pi \dfrac{\sqrt{2n}}{{3}}\right)$$

A good reference is https://en.wikipedia.org/wiki/Partition_(number_theory)