From the multinomial theorem the following holds
$$ \sum_{k_1 + k_2 + \ldots + k_m=n} {n \choose k_1, k_2, \ldots, k_m} = m^n $$
I have the following sum $$ \sum_{\substack{k_1 + k_2 + \ldots + k_m &=& B \\ g_1 + g_2 + \ldots + g_m &=& W}} {B \choose k_1, k_2, \ldots k_m} \cdot {W \choose g_1, g_2, \ldots, g_m} $$ where $B + W = n$, and $\forall_i: g_i = x - k_i$, where $x = \frac{n}{m}$, and $m\mid n$ and $k_i \geq 1$
Can this sum be simplified like the original one?
We assume positive integers $m,n\geq 2$. Since $m\mid n$ we know that $n=Nm$ is a multiple of $m$ with $N\geq 1$. This way we can write $W=Nm-B$ and $x=N$. We show by induction of $m$ that for all $N\geq 1, m\leq B\leq Nm$ the following is valid:
\begin{align*} \color{blue}{\sum_{\substack{k_1+\cdots+k_m=B\\k_1,\ldots,k_m\geq 0}} \binom{B}{k_1,\ldots,k_m}\binom{mN-B}{N-k_1,\ldots,N-k_m}=\prod_{q=2}^m\binom{qN}{N}}\tag{1} \end{align*}
Base step: $m=2$
\begin{align*} \color{blue}{\sum_{{k_1+k_2=B}\atop{k_1,k_2\geq 0}}} &\color{blue}{\binom{B}{k_1,k_2}\binom{2N-B}{N-k_1,N-k_2}}\\ &=\sum_{k_1=0}^B\binom{B}{k_1}\binom{2N-B}{N-k_1}\tag{2}\\ &=\sum_{k_1=0}^B\binom{B}{k_1}[z^{N-k_1}](1+z)^{2N-B}\tag{3}\\ &=[z^N](1+z)^{2N-B}\sum_{k_1=0}^B\binom{B}{k_1}z^{k_1}\tag{4}\\ &=[z^N](1+z)^{2N-B}(1+z)^{B}\\ &=[z^N](1+z)^{2N}\\ &\,\,\color{blue}{=\binom{2N}{N}}\tag{5} \end{align*} in accordance with (1).
Comment:
In (2) we write the multinomial as binomial coefficients and substitute $k_2 = B-k_1$.
In (3) we use the coefficient of operator $[z^N]$ to denote the coefficient of $z^N$ of a series.
In (4) we apply the rule $[z^{p-q}]A(z)=[z^p]z^qA(z)$.
In (5) we select the coefficient of $z^{N}$.
Induction hypothesis: $m=M$
We assume the validity of \begin{align*} \color{blue}{\sum_{{k_1+\cdots+k_M=B}\atop{k_1,\ldots,k_M\geq 0}} \binom{B}{k_1,\ldots,k_M}\binom{MN-B}{N-k_1,\ldots,N-k_M}=\prod_{q=2}^M\binom{qN}{N}}\tag{6} \end{align*}
Induction step: $m=M+1$
We have to show \begin{align*} \color{blue}{\sum_{\substack{k_1+\cdots+k_{M+1}=B\\k_1,\ldots,k_{M+1}\geq 0}} \binom{B}{k_1,\ldots,k_{M+1}}\binom{(M+1)N-B}{N-k_1,\ldots,N-k_{M+1}}=\prod_{q=2}^{M+1}\binom{qN}{N}}\tag{7} \end{align*}
We obtain for $M\geq 2$: \begin{align*} &\color{blue}{\sum_{\substack{k_1+\cdots+k_{M+1}=B\\k_1,\ldots,k_{M+1}\geq 0}}}\color{blue}{\binom{B}{k_1,\ldots,k_{M+1}}\binom{(M+1)N-B}{N-k_1,\ldots,n-k_{M+1}}}\\ &\qquad=\sum_{{k_1+\cdots+k_{M+1}=B}\atop{k_1,\ldots,k_{M+1}\geq 0}}\frac{B!}{k_1!\cdots k_{M+1}!}\,\frac{\left((M+1)N-B\right)!}{\left(N-k_1\right)!\cdots\left(N-k_{M+1}\right)!}\tag{8}\\ &\qquad=\sum_{k_{M+1}=0}^B\frac{B!}{k_{M+1}!\left(B-k_{M+1}\right)!}\, \frac{\left((M+1)N-B\right)!}{\left(N-k_{M+1}\right)!\left(MN-\left(B-k_{M+1}\right)\right)!}\\ &\qquad\quad\cdot\sum_{{k_1+\cdots+k_M=B-k_{M+1}}\atop{k_1,\ldots,k_{M}\geq 0}}\frac{\left(B-k_{M+1}\right)!}{k_1!\cdots k_{M}!}\,\frac{\left(MN-\left(B-k_{M+1}\right)\right)!}{\left(N-k_1\right)!\cdots\left(N-k_{M}\right)!}\tag{9}\\ &\qquad=\sum_{k_{M+1}=0}^B\binom{B}{k_{M+1}}\, \binom{(M+1)N-B}{N-k_{M+1}}\\ &\qquad\quad\cdot\sum_{{k_1+\cdots+k_M=B-k_{M+1}}\atop{k_1,\ldots,k_{M}\geq 0}} \binom{B-k_{M+1}}{k_1,\ldots, k_{M}}\,\binom{MN-\left(B-k_{M+1}\right)}{N-k_1,\ldots,N-k_{M}}\tag{10}\\ &\qquad=\prod_{q=2}^M\binom{qN}{N}\sum_{k_{M+1}=0}^B\binom{B}{k_{M+1}} [z^{N-k_{M+1}}](1+z)^{(M+1)N-B}\tag{11}\\ &\qquad=[z^{N}](1+z)^{(M+1)N-B}\prod_{q=2}^M\binom{qN}{N}\sum_{k_{M+1}=0}^B\binom{B}{k_{M+1}}z^{k_{M+1}}\\ &\qquad=[z^{N}](1+z)^{(M+1)N-B}\prod_{q=2}^M\binom{qN}{N}(1+z)^B\\ &\qquad=[z^{N}](1+z)^{(M+1)N}\prod_{q=2}^M\binom{qN}{N}\\ &\qquad=\binom{(M+1)N}{N}\prod_{q=2}^M\binom{qN}{N}\\ &\qquad\,\,\color{blue}{=\prod_{q=2}^{M+1}\binom{qN}{N}} \end{align*} in accordance with (7) and the claim follows.
Comment:
In (8) we write the multinomial coefficients using factorials.
In (9) we separate the summation of the summand $k_{M+1}$ as preparation to apply the induction hypothesis. We expand with $\left(B-k_{M+1}\right)!$ and $\left(MN-\left(B-k_{M+1}\right)\right)!$ to write the expression with binomial and multinomial coefficients.
In (10) we write the expression using binomial and multinomial coefficients.
In (11) we apply the induction hypothesis. We also use the coefficient of operator of $[z^N]$ to denote the coefficient of $z^N$ of a series.
In the following lines we use the same techniques as in the base step.
