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The literature defines multinomial coefficients (or extended bnomial coefficients) as

$$ \binom{n}{r_1,r_2,\cdots,r_l} = \frac{n!}{r_1!r_2!\cdots r_l!}$$ where $$ r_1+r_2+\cdots+r_l = n$$

Which is the coefficient of $x_1^{r_1}x_2^{r_2}\cdots x_l^{r_l}$ in expansion of $(x_1+x_2+\cdots+x_l)^n$.

Question: How do we call the coefficients of $x^r$ in expansion of $(x+x^2+\cdots+x^l)^n$. Is it bad to name them multinomial coefficients?

In case of binomial coefficients the two are same in expansion of $(x_1+x_2)^n$ and the expansion of $(x+x^2)^n$. Obviously the two are different for multinomial expansion.

Example: As $(x_1+x_2+x_3)^2= x_1^2+2 x_2 x_1+2 x_3 x_1+x_2^2+x_3^2+2 x_2 x_3$. The coefficients are $\{ 1,2,2,1,1,2 \}$. Are these MULTI-NOMIAL?

$ As (x+x^2+x^3)^2=x^2 + 2 x^3 + 3 x^4 + 2 x^5 + x^6$. The coefficients are $\{1,2,3,2,1\}$. Are these too the MULTI-NOMIAL?

I prefer different terminologies to avoid confusion. Therefore want to know how these are distinguished in existing literature.

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  • $\begingroup$ they are already called multinomial coefficients; a simple preliminary google search would've confirmed that... $\endgroup$
    – obataku
    Jul 7 '15 at 23:49
  • $\begingroup$ @oldrinb I googled but could not find a link differentiating the two. As the coefficients are entirely different it would be confusing to say multinomial in both the cases. $\endgroup$
    – Abu Bakar
    Jul 8 '15 at 2:09
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You can approach this with powerseries. The coefficient of $x^k$ in $(1+x+\cdots+x^n)^\ell$ is the way of writing $k$ as an ordered sum of elements in $S=\{0,1,\ldots,n\}$ of length $\ell$. Now note that $1+x+\cdots+x^n=\dfrac{1-x^n}{1-x}$ so that we want the coefficient of $x^k$ in $$f(x)=\frac{1}{(1-x)^\ell}(1-x^n)^\ell$$

This is thus obtained by writing out the product of the two powerseries involved, where $$\tag 1(1-x)^{-\ell}=\sum_{k\geqslant 0} \binom{\ell+k-1}{k}x^k\\(1-x^n)^{\ell}=\sum_{k\geqslant 0} \binom {\ell}k (-1)^k x^{kn}$$

Recall that $$\binom{-n}k=\frac{-n(-n-1)\cdots (-n-k+1)}{k!}=(-1)^k\frac{(n+k-1)\cdots (n+1)n}{k!}=(-1)^k\binom{n+k-1}{k}$$

which gives $(1)$. Combinatorially, the coefficient of $x^k$ in the above counts the number of ordered tuples of length $\ell$ that are solutions to $x_1+\cdots+x_\ell=k$ with entries in $\{0,\ldots,n\}$.

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  • $\begingroup$ Thank you for making the derivation easier. In this post I am looking for the names for two kind of polynomial coefficients. $\endgroup$
    – Abu Bakar
    Jul 8 '15 at 2:49
  • $\begingroup$ Pedro Tamaroff ,i think you were the one, anyway seems you have a slight error in the beginning as you have $1+x+x^2+...+x^n=\frac{1-x^n}{1-x}$ but rhs should be $\frac{1-x^{n+1}}{1-x}$ $\endgroup$
    – user158293
    Nov 2 '20 at 15:28
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$$r_1=r\\r_1+r_2+r_3+...+r_l=n\\r+r_2+r_3+...+r_l=n\\r_2+r_3+...+r_l=n-r$$so $$\binom{n-r}{r_2,r_3,...,r_l}=\frac{(n-r)!}{r_2!r_3!...r_l!}$$ it is coefficient of $x^r$

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  • $\begingroup$ $qs^n$ changed ?! $\endgroup$
    – Khosrotash
    Jul 7 '15 at 23:58
  • $\begingroup$ There was an evident typo in the question. $\endgroup$
    – Pedro Tamaroff
    Jul 8 '15 at 0:05
  • $\begingroup$ I am afraid it is not obvious to me. Say I need the coefficient of $x^3$ in expansion of $(1+x+x^2)^2$. $\endgroup$
    – Abu Bakar
    Jul 8 '15 at 2:55
  • $\begingroup$ @Galaxy Using the combinatorial interpretation, you want ways of writing $3$ as $x+y$ where $x,y$ are in $\{0,1,2\}$. This has the solutions $(0,3),(3,0),(1,2),(2,1)$ and no more, so the coefficient is $4$. $\endgroup$
    – Pedro Tamaroff
    Jul 8 '15 at 3:10

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