For $n$ urns we put $n+5$ balls in them. What's the probability for no urn being empty if the balls are all the same & if the balls are all distinct? I don't get why there should be a different result depending if the balls are all of the same color or if all balls have a different color.
What is meant by "For $n$ urns we put $n+5$ balls in them" is that we got $8$ balls for $3$ urns e.g.
 A: For your case of $8$ balls in $3$ urns then 


*

*unconstrained there are $3^8=6561$ possibilities, and these can reasonably be seen as equally probable 

*but $3$ of the $6561$ have all the balls in one urn and $762$ have all the balls in exactly two urns, 

*leaving $5796$ possible patterns with no urn empty  and a probability of $\frac{5796}{6561} \approx 0.8834$
In reality the balls are all physically distinct even if they are the same colour, and that is the way to go if you want to try this experiment. But in some people's minds there is a distribution which ignores such distinctions
Their calculation would be that if there are $8$ indistinguishable balls than $3$ distinguishable urns, then a stars and bars calculation would suggest ${10 \choose 2}=45$ possible patterns of which $3$ having all the balls in one urn and $21$ having all the balls in exactly two urns, leaving $21$ possible patterns with no urn empty and suggesting a probability of $\frac{21}{45} \approx 0.4667$
But I would argue that in the second case the patterns considered are not equally probable; in a reasonable interpretation, putting all eight balls in the first urn has a probability of $\frac{1}{6561}$ not $\frac{1}{45}$, no matter what colour the balls are.  Allowing for this will lead back to the first result
