Simplify a sum with a product and multinomial coefficient This is a follow-up question to my previous post, where I've got a great help! (I created a new one to avoid editing the original one).
Can the following sum be simplified?
$$
 \sum_{\substack{k_1 + k_2 + \ldots + k_m &=& B \\ g_1 + g_2 + \ldots + g_m &=& W}} \left(\prod_{i=1}^{m}k_i\right) \cdot {B \choose k_1, k_2, \ldots k_m} \cdot {W \choose g_1, g_2, \ldots, g_m}
$$
where $B + W = n$, and $\forall_i: g_i = x - k_i$, where $x = \frac{n}{m}$, and $m\mid n$ and $k_i \geq 1$
I tried to follow the steps suggested in my previous post, but I got stuck. For $m=2$ I've got the following
$$
\sum_{\substack{k_1 + k_2 &=& B \\ k_1, k_2 &\geq& 0}} k_1 \cdot k_2 \cdot {B \choose k_1, k_2} \cdot {2N - B \choose N-k_1, N-k_2} \\ = \sum_{\substack{k_1 + k_2 &=& B \\ k_1, k_2 &\geq& 0}} k_2 \cdot {B \choose k_1-1} \cdot {2N - B \choose N-k_1, N-k_2} \\
= \sum_{k_1 = 0}^B \left(B-k_1 \right) \cdot {B \choose k_1-1} \cdot {2N - B \choose N-k_1} \\ 
$$
From the previous post I know how to handle the binomial coefficients, but I'm not sure what can I do now with the $B-k_1$ factor inside the sum?
 A: We have
$$
S = \sum\limits_{\left\{ {\matrix{
   {k_{\,1}  + k_{\,2}  + \, \ldots  + \,k_{\,m}  = B}  \cr 
   {g_{\,1}  + g_{\,2}  + \, \ldots  + \,g_{\,m}  = W}  \cr 
 } } \right.} {\left( {\prod\limits_{i = 1}^m {k_{\,i} } } \right)
\left( \matrix{ B \cr  k_{\,1} ,\,k_{\,2} ,\, \ldots ,\,k_{\,m}  \cr}  \right)
\left( \matrix{W \cr g_{\,1} ,\,g_{\,2} ,\, \ldots ,\,g_{\,m}  \cr}  \right)} 
$$
plus the added conditions
$$
\left\{ \matrix{
  B + W = n \hfill \cr 
  g_{\,i}  = x - k_{\,i}  \hfill \cr 
  1 \le k_{\,i}  \hfill \cr}  \right.
$$
Leave apart for the moment the fact that $x=n/m, \; m | n$.
Because of the product, we can consider $0 \le ki$, and for the multinomial not to be null it shall be
$0 \le g_i$.
Thus  the set of conditions becomes (if I understood them properly)
$$
\eqalign{
  & \left\{ {\matrix{
   {0\, \le \,k_{\,i} ,\;g_{\,i} \, \le \,x}  \cr 
   {g_{\,i}  = x - k_{\,i} }  \cr 
   {k_{\,1}  + k_{\,2}  + \, \ldots  + \,k_{\,m}  = B}  \cr 
   {g_{\,1}  + g_{\,2}  + \, \ldots  + \,g_{\,m}  = W}  \cr 
 } } \right.\quad  \Rightarrow   \cr 
  &  \Rightarrow \quad \left\{ {\matrix{
   {0\, \le \,k_{\,i} ,\;g_{\,i} \, \le \,x}  \cr 
   {g_{\,i}  = x - k_{\,i} }  \cr 
   {k_{\,1}  + k_{\,2}  + \, \ldots  + \,k_{\,m}  = B}  \cr 
   {\left( {x - k_{\,1} } \right) + \left( {x - k_{\,2} } \right)
 + \, \ldots  + \,\left( {x - k_{\,m} } \right) = W}  \cr 
 } } \right.\quad  \Rightarrow   \cr 
  &  \Rightarrow \quad \left\{ {\matrix{
   {0\, \le \,k_{\,i} ,\;g_{\,i} \, \le \,x}  \cr 
   {g_{\,i}  = x - k_{\,i} }  \cr 
   {k_{\,1}  + k_{\,2}  + \, \ldots  + \,k_{\,m}  = B}  \cr 
   {W + B = mx = n}  \cr 
 } } \right. \cr} 
$$
Under these conditions the summand can be rewritten as
$$
\eqalign{
  & \left( {\prod\limits_{i = 1}^m {k_{\,i} } } \right)
\left( \matrix{ B \cr  k_{\,1} ,\,k_{\,2} ,\, \ldots ,\,k_{\,m}  \cr}  \right)
\left( \matrix{ W \cr  g_{\,1} ,\,g_{\,2} ,\, \ldots ,\,g_{\,m}  \cr}  \right) =   \cr 
  &  = \left( {\prod\limits_{i = 1}^m {k_{\,i} } } \right){{B!} \over {k_{\,1} !\,k_{\,2} !\, \cdots \,k_{\,m} !}}
{{\left( {mx - B} \right)!} \over {\left( {x - k_{\,1} } \right)!\,\left( {x - k_{\,2} } \right)!\, \cdots \,\left( {x - k_{\,m} } \right)!}} =   \cr 
  &  = {{B!\left( {mx - B} \right)!} \over {\left( {x!} \right)^{\,m} }}
\left( {\prod\limits_{i = 1}^m {k_{\,i} } } \right){{x!} \over {k_{\,1} !\,\left( {x - k_{\,1} } \right)!}}
{{x!} \over {k_{\,2} !\,\left( {x - k_{\,2} } \right)!}} \cdots {{x!} \over {k_{\,m} !\,\left( {x - k_{\,m} } \right)!}} =   \cr 
  &  = {{B!\left( {mx - B} \right)!} \over {\left( {x!} \right)^{\,m} }}
\prod\limits_{i = 1}^m {{{x\left( {x - 1} \right)^{\,\underline {\,k_{\,i}  - 1\,} } } \over {\left( {k_{\,i}  - 1} \right)!\,}}}  =   \cr 
  &  = {{B!\left( {mx - B} \right)!} \over {\left( {\left( {x - 1} \right)!} \right)^{\,m} }}
\prod\limits_{i = 1}^m {{{\left( {x - 1} \right)^{\,\underline {\,k_{\,i}  - 1\,} } } \over {\left( {k_{\,i}  - 1} \right)!\,}}}  =
 {{B!\left( {mx - B} \right)!} \over {\left( {\left( {x - 1} \right)!} \right)^{\,m} }}
\prod\limits_{i = 1}^m {\left( \matrix{ x - 1 \cr  k_{\,i}  - 1 \cr}  \right)}  \cr} 
$$
so that the sum becomes
$$
\eqalign{
  & S(B,m,x)\quad \left| {\;1\, \le mx \le B} \right.\quad  =   \cr
 &  = \sum\limits_{\left\{ {\matrix{
   {0\, \le \,k_{\,i} ,\;g_{\,i} \, \le \,x}  \cr 
   {g_{\,i}  = x - k_{\,i} }  \cr 
   {k_{\,1}  + k_{\,2}  + \, \ldots  + \,k_{\,m}  = B}  \cr 
   {g_{\,1}  + g_{\,2}  + \, \ldots  + \,g_{\,m}  = W}  \cr 
 } } \right.}
 {\left( {\prod\limits_{i = 1}^m {k_{\,i} } } \right)
\left( \matrix{B \cr k_{\,1} ,\,k_{\,2} ,\, \ldots ,\,k_{\,m}  \cr}  \right)
\left( \matrix{ W \cr g_{\,1} ,\,g_{\,2} ,\, \ldots ,\,g_{\,m}  \cr}  \right)}  =   \cr 
  &  = {{B!\left( {mx - B} \right)!} \over {\left( {\left( {x - 1} \right)!} \right)^{\,m} }}
\sum\limits_{\left\{ {\matrix{
   {\left( {1\, \le } \right)\,k_{\,i} \,\left( { \le \,x} \right)}  \cr 
   {k_{\,1}  + k_{\,2}  + \, \ldots  + \,k_{\,m}  = B}  \cr 
 } } \right.}
 {\prod\limits_{i = 1}^m {\left( \matrix{ x - 1 \cr  k_{\,i}  - 1 \cr}  \right)} }  =   \cr 
  &  = {{B!\left( {mx - B} \right)!} \over {\left( {\left( {x - 1} \right)!} \right)^{\,m} }}
\sum\limits_{\left\{ {\matrix{ {\left( {0\, \le } \right)\,j_{\,i} \,\left( { \le \,x - 1} \right)}  \cr 
   {j_{\,1}  + j_{\,2}  + \, \ldots  + \,j_{\,m}  = B - m}  \cr 
 } } \right.}
 {\prod\limits_{i = 1}^m {\left( \matrix{ x - 1 \cr  j_{\,i}  \cr}  \right)} }  \cr} 
$$
Now, since
$$
\eqalign{
  & \left( {\left( {1 + z} \right)^{x - 1} } \right)^{\,m}
  = \prod\limits_{i = 1}^m {\left( {\sum\limits_{\left( {0\, \le } \right)\,j_{\,i} \,\left( { \le \,x - 1} \right)}
 {\left( \matrix{ x - 1 \cr  j_{\,i}  \cr}  \right)z^{\,j_{\,i} } } } \right)}  =   \cr 
  &  = \sum\limits_{0\, \le \,s\, \le \,m\,\left( {x - 1} \right)}
 {\sum\limits_{\left\{ {\matrix{ {\left( {0\, \le } \right)\,j_{\,i} \,\left( { \le \,x - 1} \right)}  \cr 
   {j_{\,1}  + j_{\,2}  + \, \ldots  + \,j_{\,m}  = \,s}  \cr 
 } } \right.}
 {\left( {\prod\limits_{i = 1}^m {\left( \matrix{ x - 1 \cr j_{\,i}  \cr}  \right)} } \right)z^{\,s} } }  \cr} 
$$
we conclude that
$$ \bbox[lightyellow] {  
\eqalign{
  & S(B,m,x)\quad \left| {\;1\, \le mx \le B} \right.\quad  =   \cr
 & = {{B!\left( {mx - B} \right)!} \over {\left( {\left( {x - 1} \right)!} \right)^{\,m} }}
\;\;\left[ {z^{\,B - m} } \right]\left( {1 + z} \right)^{m\left( {x - 1} \right)}  =   \cr 
  &  = {{B!\left( {mx - B} \right)!} \over {\left( {\left( {x - 1} \right)!} \right)^{\,m} }}
\;\;\left( \matrix{ m\left( {x - 1} \right) \cr  B - m \cr}  \right) =   \cr 
  &  = {{B!\left( {mx - B} \right)!} \over {\left( {\left( {x - 1} \right)!} \right)^{\,m} }}
\;{{\left( {m\left( {x - 1} \right)} \right)!} \over {\left( {B - m} \right)!\left( {mx - B} \right)!}}\; =   \cr 
  &  = {{B!} \over {\left( {B - m} \right)!}}\;
{{\left( {m\left( {x - 1} \right)} \right)!} \over {\left( {\left( {x - 1} \right)!} \right)^{\,m} }}\;
 = \left( \matrix{ B \cr  m \cr}  \right)\;
{{m!\left( {m\left( {x - 1} \right)} \right)!} \over {\left( {\left( {x - 1} \right)!} \right)^{\,m} }} \cr} 
}$$
--- in reply to your comment  ---
Consider the following simple case
$$
\eqalign{
  & \left( {1 + z} \right)^a \left( {1 + z} \right)^a
  = \left( {\sum\limits_{k_1 } {\left( \matrix{ a \cr  k_1  \cr}  \right)z^{k_1 } } } \right)
\left( {\sum\limits_{k_2 } {\left( \matrix{ a \cr   k_2  \cr}  \right)z^{k_2 } } } \right) =   \cr 
  &  = \prod\limits_{i = 1}^2 {\left( {\sum\limits_{k_i }
 {\left( \matrix{a \cr k_i  \cr}  \right)z^{k_i } } } \right)}  =   \cr 
  &  = \sum\limits_{k_1 } {\sum\limits_{k_2 }
 {\left( \matrix{ a \cr  k_1  \cr}  \right)\left( \matrix{ a \cr k_2  \cr}  \right)z^{k_1  + k_2 } } }
  = \sum\limits_s {\sum\limits_{k_1  + k_2  = s}
 {\left( \matrix{a \cr k_1  \cr}  \right)\left( \matrix{a \cr   k_2  \cr}  \right)z^s } }  =   \cr 
  &  = \sum\limits_s {\left( {\sum\limits_{k_1  + k_2  = s} {\left( {\prod\limits_{i = 1}^2
 {\left( \matrix{ a \cr  k_i  \cr}  \right)} } \right)} } \right)z^s }
  = \sum\limits_s {\left( \matrix{ 2a \cr   s \cr}  \right)z^s }  \cr} 
$$
which implies
$$
\sum\limits_{k_1  + k_2  = s} {\left( {\prod\limits_{i = 1}^2
 {\left( \matrix{a \cr  k_i  \cr}  \right)} } \right)}
  = \left( \matrix{ 2a \cr   s \cr}  \right)
$$
and which is just the Vandermonde convolution.
