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The binomial coefficients represent the coefficients on the terms in the expansion of $(x+y)^n$, but they can also be interpreted as choosing a subset of items from a set while disregarding the order of choice and disallowing repeated choices.

Multinomial coefficients are the coefficients in the expansion of $(x1+x2+...+x_m)^n$, but I have not yet been able to think of a "choose" related interpretation of multinomial coefficient in the same vein as the interpretation of binomial coefficients.

The only interpretation I know of is that of counting the number of strings with repeated letters, e.g. the number of rearrangements of MISSISSIPPI is the multinomial $\frac{11!}{1!4!4!2!}$. I suppose binomials can also be thought of as counting the number of strings with repeated letters, made out of only two letters, e.g. $6 \choose 4$ is the number of rearrangements of AAAABB.

But is there an interesting interpretation of multinomials in terms of forming subsets or otherwise "choosing" items in some way? I know it's not choosing elements disregarding order but allowing repeated choices; that problem is solved by multichoose aka stars and bars. So what could the interpretation of multinomials be?

xojfqa
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    The multinomial coefficient $\binom{N}{n_1,n_2,\dots,n_k}$ where $n_1+n_2+\dots+n_k=N$ is equal to the number of ways to take $N$ distinct elements and partition them into $k$ labelled parts such that $n_1$ elements are placed into the first part, $n_2$ into the second part and so on. For the example of MISSISSIPPI arrangements, the elements are the positions in the word, the parts are labelled M,I,S,P respectively. For the part labelled S in the original arrangement that is the part made up of the third, fourth, sixth, and seventh positions. – JMoravitz Aug 06 '22 at 15:22
  • In the way you want the definition, here is one which I had picked up from somewhere: "The multinomial coefficient can be defined as counting the number of ways of splitting a set of n elements into an ordered sequence of k disjoint subsets, each subset being ≥0" – true blue anil Aug 06 '22 at 19:52

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You can write a multinomial coefficient as a product of binomial coefficients. If $n,m$ are nonengative integers, and $k_1,\dots,k_m$ are nonnegative integers summing to $n$, then $$ \binom{n}{k_1,k_2,\dots,k_m}=\binom{n}{k_1} \binom{n-k_1}{k_2}\binom{n-k_1-k_2}{k_3}\cdots \binom {n-k_1-\dots-k_{m-2}}{k_{m-1}}. $$ This leads to the following combinatorial interpretation; the multinomial coefficient is the number of ways to choose $k_1$ objects from a pool of $n$, then to choose $k_2$ objects from the remaining pool of $n-k_1$ objects, then choose $k_3$ objects from the remaining pool of $n-k_1-k_2$ objects, and so on. So, you are not choosing a single subset, but rather $m-1$ disjoint subsets.

Equivalently, this is the number of ways to take $n$ distinct objects, and place them each into one of $m$ distinct boxes, so that box number $i$ gets exactly $k_i$ objects for each $i\in \{1,\dots,m\}$.

Mike Earnest
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  • The wording distinct objects into distinct boxes confuses me. Isn't it that the objects get labeled by the box they are put in rather than that each object is actually distinct. Suppose I randomly throw indistinguishable balls into distinguishable boxes, the multinomial coefficient will apply for whatever pattern is formed, won't it ? Even when I simply divide n people into m labeled teams, aren't the people treated as faceless (unless other criteria are involved) ? – true blue anil Aug 06 '22 at 18:14
  • @trueblueanil for the contexts in which multinomial coefficients are used... they are absolutely distinct objects. "Suppose I randomly throw indistinguishable balls into distinguishable boxes" If you are doing it in sequence, one ball at a time... then the balls are distinguishable based on the timing in which they were thrown. Otherwise if they truly are indistinguishable (to our mortal eyes) then they would not be counted in this way but rather with stars-and-bars. The punchline is that if we were to try to ask probability questions about this scenario, then stars-and-bars is wrong. – JMoravitz Aug 06 '22 at 18:29
  • @trueblueanil Can you give an explicit choice of integers $n,m,k_1,\dots,k_m$, for which $\binom{n}{k_1,\dots,k_m}$ is not equal to the number of ways to place $n$ distinct objects into $m$ distinct boxes such that the $i^{th}$ box gets $k_i$ objects? If not, then I do not understand your confusion. – Mike Earnest Aug 06 '22 at 18:30
  • @trueblueanil Despite whether or not to our mortal eyes we can distinguish the objects... "water has memory." Each of the objects occupies a different position in space and so we do have a way of distinguishing them based on that. We can stick postit notes on them or color them... or we can examine them very closely and find that in fact this ball is 0.000001 grams heavier than the other... or spot a serial number... etc... The outcomes counted by stars and bars are not equally likely occurrences. – JMoravitz Aug 06 '22 at 18:32
  • @JMoravitz: Ok, if you say that the sequence, etc also confer distinctness, I understand. – true blue anil Aug 06 '22 at 18:50
  • Thank you, this is very intuitive. Since in general the number of ways to put $n$ distinct objects into $m$ distinct boxes where order matters is $m^n$, and the number of ways to do so such that each box gets a specific number of objects must be strictly less than that, can we use this to conclude that $\frac{n!}{k_1!...k_m!} < m^n$ for all $m, n, k_1, ..., k_m$? – xojfqa Aug 08 '22 at 22:09
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    Indeed you can! This generalizes the inequality $\frac{n!}{k!(n-k)!}<2^n$. – Mike Earnest Aug 08 '22 at 22:16