Identity for a product of choose functions Given non-negative integers $k$, $m$, $n$, where $n$ $\geq$ $m$ and $n$ $\equiv$ $m$ (mod $2$), can you show that:
$${n + k \choose m} {\frac{n - m}{2} + k \choose k} = \sum_{∀i, m_i \equiv n_i (mod 2)} \left[\prod_{i=0}^{k} {n_i \choose m_i}\right]$$
where $\sum_{i=0}^{k} m_i$ = $m$, $\sum_{i=0}^{k} n_i$ = $n$, and $m_i$, $n_i$ $\geq$ $0$ and $m_i$ $\leq$ $n_i$, $i$ = $0$, $k$.
The following is context on how I arrived to post this question. I posted this question after empirically verifying (up to D = 20) an answer essentially in the form of the left hand side of the equation above. Just recently, I set about to logically prove this result. I doing so, I wound up with an expression in the form of the right hand side of the equation. Thus, the above identity lies at the heart of the $CQHRL_D$ face counting question.
As to my progress on deriving this identity, I put myself at the starting line alongside others who are much more knowledgeable and capable; however I can count interest in the problem in my favor.
 A: We will establish the identity by counting the elements of the same set two different ways.
Begin by considering the trinomial coefficient for ($m$, $n$ - $m$, $k$); i.e., the coefficient of the $a$m$b$n-m$c$k term in the expansion of ($a$ + $b$ + $c$)n+k.
This coefficient can be calculated by counting the number of corresponding terms in the full expansion of ($a$ + $b$ + $c$)n+k. The number of combinations of positions for the $m$ $a$ factors is ${n+k \choose m}$. For each of these, there are ${n-m+k \choose k}$ ways to arrange the $b$ and $c$ factors, for a total of ${n+k \choose m}$${n-m+k \choose k}$.
Also, note that in each of these terms, there are $k+1$ stretches of only $a$ & $b$ factors (possibly including the empty set); i.e., before the first $c$ factor, between each pair of subsequent $c$ factors, and after the last $c$ factor. Each combination of (positioning of the $k$ $c$ factors, partition of the $m$ $a$ factors into $m_i$ $a$ factors in the $i$th stretch) corresponds to $\prod_{i=0}^{k} {n_i \choose m_i}$ terms in the full expansion, where the $i$th stretch of only $a$ & $b$ factors has $m_i$ $a$'s and ($n_i$ -$m_i$) $b$'s; hence $\sum_{i=0}^{k} m_i$ = $m$ and $\sum_{i=0}^{k} n_i$ = $n$. (Note that $n_i$ - the length of the $i$th only $a$ & $b$ stretch - is determined by the positioning of the $c$'s.) The total coefficient is obtained by summing the contributions over all such combinations of positionings of the $k$ $c$ factors and partitions of the $m$ $a$ factors, or $\sum \left[\prod_{i=0}^{k} {n_i \choose m_i}\right]$; where 
$\sum_{i=0}^{k} m_i$ = $m$ and $\sum_{i=0}^{k} n_i$ = $n$.
Now, we wish to count the elements of a subset of the terms described above - only those terms where there are an even number of $b$ factors before the first $c$ factor, between each pair of subsequent $c$ factors, and after the last $c$ factor. (As we are given that $n$ $\equiv$ $m$ (mod $2$), $n-m$ is even, so we know this is possible.) Again, the number of combinations of positions for the $m$ $a$ factors is ${n+k \choose m}$. For the remaining factors, consider $bb$ as a unit, as we know $b$'s must occur in pairs among the $c$'s. So the number of eligible arrangements of $b$'s and $c$'s is ${\frac{n - m}{2} + k \choose k}$, for a total count of ${n+k \choose m}$${\frac{n - m}{2} + k \choose k}$.
Counting the other way, each eligible combination of (positioning of the $k$ $c$ factors, partition of the $m$ $a$ factors into $m_i$ $a$ factors in the $i$th stretch) again corresponds to $\prod_{i=0}^{k} {n_i \choose m_i}$ terms. To be eligible, ($n_i$ - $m_i$) must be even ∀$i$, which is equivalent to saying $n_i$ $\equiv$ $m_i$ (mod $2$) ∀$i$. So, the count is $$\sum_{∀i, m_i \equiv n_i (mod 2)} \left[\prod_{i=0}^{k} {n_i \choose m_i}\right]$$
A similar argument shows that given non-negative integers $k$, $m$, $n$, and $q$ where $n$ $\geq$ $m$, $q$ > $0$, and $n$ $\equiv$ $m$ (mod $q$),
$${n + k \choose m} {\frac{n - m}{q} + k \choose k} = \sum_{∀i, m_i \equiv n_i (mod  q)} \left[\prod_{i=0}^{k} {n_i \choose m_i}\right]$$
