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The multinomial theorem:

$(x_1 + x_2 + \cdots + x_m)^n = \sum_{k_1+k_2+\cdots +k_m = n} {n \choose k_1, k_2, \ldots, k_m} x_1^{k_1} x_2^{k_2} \cdots x_m^{k_m}$

There are lots of dots here, and to be fancy and rigorous (one could rise the inevitable question, What do the dots mean? We have never had them defined!) we should find a notation without the dots. Let's try:

$\left(\sum_{i=1}^nx_i\right)^n=\sum_{k_i,\ \sum_{i=1}^mk_i=n}{n \choose k_1, k_2, \ldots, k_m}\prod_{i=1}^mx_i^{k_i}$

Much less readable, but who cares, its rigorous, contains no undefined notations, and has no dots!

Ugh. It does have dots. Here: $n \choose k_1, k_2, \ldots, k_m$ Unacceptable!

Are these ugly dots the only way to denote the multinomial coefficient? Is there any notation that would avoid the dots?

EDIT: Ah, and now I can see one more problem with rigorousness. Here: $\sum_{k_i,\ \sum_{i=1}^mk_i=n}$ We give conditions on $k_i$ where $1\leq i\leq m$, but there are no conditions on $k_i$ values where $i$ is outside of these bounds! A nitpicker could say, this is unacceptable, what is the value of $k_{n+1}$? How to denote in the above formula that we forbid any values of $i$ other than these: $1\leq i\leq m$?

gaazkam
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    I think you're fighting a losing battle here. The dots are incredibly useful for representing an undefined number of terms. The barrier to understanding what they mean is extremely minimal, as well. I would suggest redirecting your energies to something better. – FullofDill Dec 04 '17 at 17:47
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    $k_{n+1}$ is undefined, but that is unproblematic, because it's unused as well. – Arthur Dec 04 '17 at 18:00
  • Pretty not sure if this help. https://arxiv.org/abs/1105.4135 – Eric Dec 04 '17 at 18:02
  • @FullofDill : Ikr. I'm asking this question out of curiosity only, not because I think it has any practical value. I am not ideologically opposed to dots, only out of curiosity I wonder if it is possible to eliminate them. – gaazkam Dec 04 '17 at 18:15
  • @Arthur The problem is, the bounds for $i$ are not present in the formula, so a nitpicker could say we don't know $k_{n+1}$ is unused. We only know what conditions the values of $k_1,\cdots,k_n$ have to fulfill, but we don't know anything about $k_{n+1},\cdots,k_{\infty}$. – gaazkam Dec 04 '17 at 18:19
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    Define $\binom{n}{k}$, where $k $ is $m$ dimensional. Define $x^k$ as appropriate and define $|k|=\sum_i k_i$. Then write $\sum_{|k|=n} \binom{n}{k} x^k$. This will reduce global dotting. – copper.hat Dec 04 '17 at 18:56

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