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From A First Course in Probability, 8th Ed - Sheldon Ross.

My solution:

$$ \frac{{n \choose m} m! (N-1)^{(n-m)}}{N^n}$$

My reasoning is that if we take the denominator to be $N^n$, it seems we're taking an ordered approach, and so the numerator should follow suit.

If we choose m of the n balls to be in the first compartment, then we have ${n \choose m} m!$ ordered choices.

The other balls can be in any other compartment. There are N-1 choices of comparment and (n-m) balls left, so that's where the other term comes from.

However, the solution is:

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    $\begingroup$ The book solution is correct. What matters is which balls are in which compartment. The question does not state that there is a way to order balls within a compartment. If balls had to be ordered within each compartment, then you cannot just order them in the first compartment and leave others. You will have to order them in each compartment then. Note that $N^n$ puts $n$ balls into $N$ compartments but it does not order them within each compartment. $\endgroup$
    – Math Lover
    Commented Apr 14, 2021 at 10:09
  • $\begingroup$ I started to write down the all the events for n = 3, N = 3 and m = 2 and I get your conclusion! Thank you. Just one other thing, why is the other term (n-1)^(n-m) not (N-1)^(n-m) as our choice of compartments has gone down by 1? $\endgroup$
    – threelinewhip
    Commented Apr 14, 2021 at 10:16
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    $\begingroup$ @MathLover The book solution is not correct. It should be $(N-1)$ instead of $(n-1)$. So there is still a typo in the book. $\endgroup$
    – user
    Commented Apr 14, 2021 at 10:38
  • $\begingroup$ @user yes I did not notice $N$ vs $n$ in the book answer. Nonetheless, my point to OP was on multiplication by $m!$. $\endgroup$
    – Math Lover
    Commented Apr 14, 2021 at 10:41
  • $\begingroup$ @threelinewhip on that you are right. There is a typo in the book - it should be $(N-1)$ and not $(n-1)$ $\endgroup$
    – Math Lover
    Commented Apr 14, 2021 at 10:42

1 Answer 1

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The book should have said $${{n \choose m} (N-1)^{n-m}}/{N^n}$$ with $(N-1)^{n-m}$ rather than $(n-1)^{n-m}$.

This is a simple binomial distribution calculation ${n \choose m}\left(\frac1N\right)^m\left(\frac{N-1}N\right)^{n-m}$, with each ball independently having probability $\frac{1}{N}$ of going into the first compartment and $\frac{N-1}{N}$ of going into another compartment. You do not have to multiply by $m!$ because you have already taken into account the natural ordering of the balls: the ${n \choose m}$ term counts the different ways which $m$ of the $n$ balls go into the first compartment.

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  • $\begingroup$ Sound. Self-teaching undergrad probability and hadn't (re-)looked at distributions yet, so this was a nice reminder of how to apply the binomial distribution. $\endgroup$
    – threelinewhip
    Commented Apr 14, 2021 at 10:34

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