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This question was taken from Mathematical Statistics and Application $7$th edition page $85$ , question $2.181$

Suppose that n indistinguishable balls are to be arranged in N distinguishable boxes so that each distinguishable arrangement is equally likely. If n ≥ N, show that the probability no box will be empty is given by $$\frac{\binom{n-1}{N-1}}{\binom{N+n-1}{N-1}}$$

It has also an answer such that : Suppose that $n$ indistinguishable balls are arranged in N distinguishable boxes

However , i have a problem such that as far as i see probability questions in M.S.E , it is said that when we deal with probability , it is not recommended using combination with repetition formula. The best method for indistinguishable balls into distinguishable urns is that thinking it like distinguishable balls into distinguishable urns , because when we select any indistinguishable balls , it become distinguishable. So , i want you to enlighten me. The solution of book is given and it makes sense , but many experts do not recommend to use conbination with repetition. Which approach is correct ?

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The distinction comes from the phrase "so that each distinguishable arrangement is equally likely" in the problem statement:

Suppose that n indistinguishable balls are to be arranged in N distinguishable boxes so that each distinguishable arrangement is equally likely. If n ≥ N, show that the probability no box will be empty is given by

It is thanks to this phrase being included in the problem statement that we may assume the $\binom{N+n-1}{N-1}$ arrangements are equally likely... because that is precisely what the phrase tells us to do. Without this phrase being included, such an assumption is invalid and the assumption of each ball being placed uniformly and independently at random is far more standard which is what leads to the sample space of size $N^n$ equally likely outcomes.

Remember, for an event $A$ and a sample space $S$ we may use $\Pr(A)=\dfrac{|A|}{|S|}$ as a formula only when $S$ is known to consist of equally likely outcomes. There are two outcomes to playing the lottery, you win or you lose. You don't win with probability $\frac{1}{2}$ however. For the standard interpretation with balls being separately placed uniformly and independently at random into boxes in sequence, even if we choose to "forget which ball was which" after having placed them... those $\binom{N+n-1}{N-1}$ outcomes counted by stars and bars are not equally likely to have occurred. To see this, don't forget which ball was which and you see the $N^n$ outcomes are equally likely, and you see that some outcomes are more prevalent than others (for example one ball in box2 and the rest in box1 is $n$ times more likely than all balls in box1).

JMoravitz
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  • So , the main point is "so that each distinguishable arrangement is equally likely" , because of it is said , we were able to use combination with repetiton , but if it not stated , we solved the question like distinguishable objects into distinguishable boxes , even if the balls are indistinguishable , right ? –  Oct 22 '21 at 12:46
  • Yes, that's right. It all comes down to what assumptions we make about the random process that delivers the balls into the boxes... Technically, without any information about the random processes involved we are stuck and have no way of approaching any problem... however if we were to make an assumption about how the balls get into the boxes in the first place, then we can proceed and the most common and most natural way is again to go through the balls in sequence, one at a time, and place them into a box uniformly and independently. – JMoravitz Oct 22 '21 at 12:50
  • Even though the balls may be indistinguishable at sight to us, they may yet be distinguished based on when they were placed into a box which affords us the ability to consider the $N^n$ equally likely outcomes separately even if many of those outcomes are indistinguishable to us after the fact. For more on this, see my answer to Why is flipping a head then a tail a different outcome than flipping a tail then a head? – JMoravitz Oct 22 '21 at 12:51
  • For curiosity , what if the balls and urns were indistinguishable , would we assume that both of them distinguishable without deling with partition –  Oct 22 '21 at 12:52
  • Usually, yes.$~$ – JMoravitz Oct 22 '21 at 12:56