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To count the ways of sorting $n$ objects into $m$ bins where there are $k_i$ objects in the $i$-th bin, we use the multinomial coefficient ${{k_1+k_2+...+k_m}\choose{k_1}}{{k_2+k_3+...+k_m}\choose{k_{2}}}... {{k_{m-1}+k_m}\choose{k_{m-1}}} =\frac{n!}{k_1!k_2!...k_m!}$, but sometimes we want to divide by $m!$ because we don't care about the order in which we fill the bins. There are $m!$ permutations of the bins, but if the bins are identical then the permutations are all equivalent, thus we calculate $\frac{n!}{m!k_1!k_2!...k_m!}$. I don’t understand why this same logic does not apply to the "balls in bins" frame of counting problems.

$n^k$ counts the ways to put $k$ distinct objects into $n$ distinct bins, so if we divide by the number of permutations of the bins, which is $n!$, then we should arrive at the case where the bins are identical. Yet the “distinct objects into identical bins” problem is solved by Stirling and Bell numbers, which look nothing like $\frac{n^k}{n!}$

Likewise, the problem of putting $k$ identical objects into $n$ distinct bins is solved with ${k+n-1}\choose{k}$ (stars and bars), so the “identical objects into identical bins” problem would be solved by $\frac{{k+n-1}\choose{k}}{n!}$. But that is clearly wrong since it potentially results in a noninteger.

Why does dividing by $n!$ not serve to control for distinctness of bins in these problems, and what then is the actual combinatorial interpretation (if any) of $\frac{n^k}{n!}$ and $\frac{{k+n-1}\choose{k}}{n!}$?

bdavidson
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  • May I have a reference of your first formula $\frac{n!}{m!k_1!k_2!...k_m!}$? This is not so intuitive to me. – Vezen BU Aug 24 '22 at 14:16
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    We could interpret $\frac {n^k}{n!}$ as: You have $k$ bins of the same $n$ items. You draw one from each bin. If you have two or more of the same items from different bins you make them distinct. But then order is unimportant. This is "combinatorally contradictory" because it makes items simultaneous both distinct and indistinct. Which accounts it not being an integer. It'd be like $\frac{10000}{16}$ is the number of $4$ digit strings where order doesn't matter but if you have multidigits you distinguish them by order but then you allow the impossible where the 2 digit appeared 3rd. – fleablood Aug 24 '22 at 16:56
  • @VezenBU first we choose $k_1$ from all elements, then choose $k_2$ from the remaining $n-k_1$, and so on. If we write out these "choose"s then terms in the denominators and numerators cancel in such a way as to leave $\frac{n!}{k_1!k_2!...k_m!}$ – bdavidson Aug 24 '22 at 18:03
  • @VezenBU let me know if that helps at all but if not I can find a resource that explains it better – bdavidson Aug 24 '22 at 18:04
  • @VezenBU Apologies, I just realized you were referring to the version of this formula which was additionally divided by $m!$, and I have in fact realized that such a division is not valid (or at least only works in limited cases, such as when pairing up elements) – bdavidson Aug 24 '22 at 18:25
  • So for example: $10^4$ then number of four digits string. Where $4387$ is distinct from $3847$. Naively (and wrong) the $\frac{10^4}{4!}$ is where $4387$ is the same as $3847$. Now if we have $407_17_2$ that to be the same as $07_247_1$ but consider $07_147_2$ to also be the same but somehow be a different way of being the same where order doesn't matter but those with repeated digit are deliberately over counted and weighted heavier. That'd be $\frac {10^4}{4!}$. But I can't think of any practical or sensible use for that concept. – fleablood Aug 24 '22 at 19:49
  • @fleablood Thanks for this, I've realized from your comments that my confusion has indeed been exactly a confusion between the ideas of ordering and distinctness. I would think that ordering objects makes them distinct, and being distinct gives them an ordering, but (if I understand your example correctly) this is not true. – bdavidson Aug 24 '22 at 20:20

1 Answers1

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Your assertion in the first paragraph is not correct.

Let's try to find a meaning for $\frac{n^k}{k!}$
Consider the simple case of putting $4$ distinct objects into $3$ distinct bins

Using the format $\,$[ Lay down pattern ]$\times$[permute pattern], we get

$4-0-0: \binom 4{4,0,0}\frac{3!}{2!} =3$

$3-1-0: \binom 4{3,1,0} 3! = 24$

$2-2-0: \binom 4{2,2,0}\frac{3!}{2!} = 18$

$2-1-1: \binom 4{2,1,1}\frac{3!}{2!} = 18 \to \boxed{63}$

What, then do you think $\Large\frac{63}{3!}$ indicates ?
I can't attribute any useful meaning to $\Large\frac{n^k}{k!}$

And the same holds good for the stars and bars version of $\binom{4+3-1}{3-1} = 15$ placements

ADDED

The punchline is that the multinomial coefficient puts distinct (=labeled) balls into distinct (=labeled) bins with matching labels, [$k_i$ into box $i$, not into any box], that is why when we want to count all ways of putting distinct objects into distinct boxes,[=$n^k$] we need to multiply two multinomial coefficients, one for the numner of possible patterns, and one for permuting each pattern, ergo $\frac{n^k}{k!}$ doesn't have a useful meaning

true blue anil
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  • There is at least some precedent for dividing the multinomial coefficient by $m!$, such as when counting pairs: https://math.stackexchange.com/questions/1425250/in-how-many-ways-can-we-pair-ourselves/1425262#1425262 . But I see now I was wrong to generalize this, as demonstrated by your example. I'm now trying to find the conditions where the reasoning given in that linked answer does apply, but I suppose I should not ask a separate question in a comment. – bdavidson Aug 24 '22 at 18:39
  • @bdavidson: Your link is an example of, say, pairing $2n$ socks of $n$ different colors, where socks within a pair are indistinguishable, and order in which pairs are placed doesn't matter – true blue anil Aug 24 '22 at 18:55