Questions tagged [multinomial-coefficients]
For questions related to multinomial coefficients, a generalization of binomial coefficients.
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Categorical distribution pmf
I am trying to understand the pmf $p(y|\theta_1,\dots,\theta_c)=\Pi_{k=1}^c\theta_k^{y_k}$ of the categorical distribution but I do not understand why there aren't any $1-\theta_k$ terms, like in the ...
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On asymptotics of certain sums of multinomial coefficients
Given positive integers $n$ and $k$, set
$$ S_{n,k}=\sum_{\substack{a_1+a_2+\dots+a_k=2n\\ a_i \in 2\mathbb{N},\,i=1,\ldots,k}}\frac{(2n)!}{a_1!a_2!\dots a_k!},
$$
where $2\mathbb{N}=\{0,2,4,\ldots\}$....
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A sum of multinomial coefficients over partitions of integer
I denote a partition of an integer $n$ by $\vec i = (i_1, i_2, \ldots)$ (with $i_1, i_2, \ldots \in \mathbb N$) and define it by
$$
\sum_{p\geq1} p i_p = n.
$$
I set
$$
|\vec i| = \sum_{p\geq1} i_p.
$$...
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How to find the coefficient of $x^3y^4z$ in $ (x+y+z)^5 (1+x+y+z)^{5}$?
First of all, I know that there is an extremely similar question from yesterday that has been closed due to Mathematics Stack Exchange guidelines, so I can't comment and find what is incorrect in my ...
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Generating Function for Modified Multinomial Coefficients
The multinomial coefficients can be used to expand expressions of the form ${\left( {{x_1} + {x_2} + {x_3} + ...} \right)^n}$ in the basis of monomial symmetric polynomials (MSP). For example,
$$\...
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Multiindex partial derivative higher order product rule, i.e. formula for $\partial^\alpha(fg)$
I want to prove the product rule for higher order partial derivatives. It is given on Wikipedia under the name "General Leibniz rule":
$$\partial^\alpha(fg)=\sum_{\beta\leq \alpha}\binom\...
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If $(1+px+x^2)^n=1+a_{1}x+a_{2} x^2+....+a_{2n}x^{2n}$, then prove that $(np-pr)a_{r}=(r+1)a_{r+1}+(r-1-2n)a_{r-1}$ for $1<r<2n$
If $(1+px+x^2)^n=1+a_{1}x+a_{2} x^2+....+a_{2n}x^{2n}$, then prove that $(np-pr)a_{r}=(r+1)a_{r+1}+(r-1-2n)a_{r-1}$ for $1<r<2n$
My try:
I tried putting $r=2$ and solved the problem and verfied ...
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Demonstrating a Binomial Identity? #2 (The exclusion)
$$
\sum_{m=1}^{\lfloor j/(k+1) \rfloor}(-1)^m\binom{n}m\binom{j-m(k+1)+n-1}{n-1}
= \sum_{m=1}^{j-k} {j-k-1 \choose m-1}{n \choose m}m
$$
(Actually $\not =$ see edit end of post)
Is there a simple ...
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Binomial identity?
$${n+k-1 \choose k}=\sum_{m=1}^{min(k,n)}{k-1 \choose m-1}{n \choose m} $$
Is there a simple way to demonstrate this equality?
Context
These are two ways of expressing the $x^k$ coefficients in $(1+x+...
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Extracting coefficients with the power series $(1-x)^{-n}$
Given polynomials of the form $$(1+x+x^2+x^3+\cdots+x^k)^n $$ We can calcualte the coefficients by writing it in the form $$(1-x^{k+1})^n \over (1-x)^n$$ and using the power series $(1-x)^{-n}$, as ...
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Combinatorial interpretation of the multinomial coefficient as a product of binomial coefficients
$$
\begin{align}& \binom{n}{k_1,k_2,\dots,k_m}\\&=\frac{n!}{k_1!k_2!\cdots k_m!}\\&=\binom{k_1}{k_1} \binom{k_1+k_2}{k_2}\cdots \binom {k_1+k_2+\cdots+k_{m}}{k_{m}}\end{align}
$$
Is there ...
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Prove $\sum_k{{a+b \choose a+k}{b+c \choose b+k}{c+a \choose c+k}(-1)^k}=\dfrac{(a+b+c)!}{a!b!c!}$ [duplicate]
How can I prove this: $$\sum_k{{a+b \choose a+k}{b+c \choose b+k}{c+a \choose c+k}(-1)^k}=\dfrac{(a+b+c)!}{a!b!c!}$$
I know I should avoid no clue questions, but really I have no idea about this one. ...
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Question: Concerning Simplifying Random Walk from 2D to 1D
I have a question that has been confusing me. For a 1D random walk in the x-direction I was told that the multinomial coefficient is given by:
$$C(N,k_x) = \frac{N!}{k_x!(N-k_x)!} \tag{1}$$
In Eq. 1, ...
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Weighted sum of specific multinomial coefficients
Let $A$ and $b$ be nonnegative integers and consider the sums
$$\sum\limits_{c=0}^{b/2}\frac{1}{4^c}\binom{A}{c,b-2c,A-b+c}$$
and
$$\sum\limits_{c=0}^{b/2}\frac{c}{4^c}\binom{A}{c,b-2c,A-b+c}.$$
I ...
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Calculating Coefficents of a single variable polynomial [duplicate]
Given:
$$
(1+x+x^2+x^3+\cdots+x^k)^n
$$
Is there a formula to calculate the coefficient of $x^a$ (where $a$ can be any integer value less than $k^n$) that's more efficient than grinding through ...
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