I'm trying to find the following maximum: $\max\_{\gamma}\sum\_{|\alpha|=q}\binom{\alpha}{\gamma}$. Here $\alpha=(\alpha\_1,\ldots, \alpha\_n),\gamma=(\gamma\_1,\ldots, \gamma\_n)$ are multi-indices. The binomial coefficient is defined as $\binom{\alpha}{\gamma}=\frac{\alpha !}{\gamma! (\alpha-\gamma)!}=\prod\_{i=1}^n \frac{\alpha\_i !}{\gamma\_i! (\alpha\_i-\gamma\_i)!}=\prod\_{i=1}^n\binom{\alpha\_i}{\gamma\_i}$. We take the usual convention that $\binom{n}{r}=0$ if $r$ goes out of range, i.e. $r<0$ or $r>n$.


This maximum is well-defined and fully determined in terms of $n$ and $q$. Could anyone help? 


Here is the solution for $n=1$. The sum $\sum\_{|\alpha|=q}\binom{\alpha}{\gamma}$ is the single term $\binom{q}{\gamma}$, so the monotonicity of the binomial distribution gives the maximum at $\binom{q}{\lfloor q/2\rfloor}$.


For $n=2$, the problem amounts to maximizing $\max\_{r,s}\sum\_{i=0}^q \binom{i}{r}\binom{q-i}{s}=\max\_{r,s}\sum\_{i=r}^{q-s} \binom{i}{r}\binom{q-i}{s}$.


For general $n$, here is my very rough estimate. The basic inequality $\binom{\alpha}{\gamma}\le \binom{|\alpha|}{|\gamma|}$ implies $\max\_\gamma \sum\_{|\alpha|=q}\binom{\alpha}{\gamma}\le \max\_{\gamma}\sum\_{|\alpha|=q}\binom{q}{|\gamma|}\le \binom{n+q-1}{n}\binom{q}{\lfloor q/2\rfloor}$. The coefficient $\binom{n+q-1}{n}$ that pops out is the number of multi-indices $\alpha$ of length $q$.


More rewrites and updates to come. Thanks to Gerhard in advance!



How do i maximize $\max_{\gamma}\sum_{|\alpha|=q}\binom{\alpha}{\gamma}$?
I claim that $$\sum\_{|\alpha|=q} \binom{\alpha}{\gamma}=\binom{n+q-1}{|\gamma|+n-1}$$ 
and so the answer is simply all $\gamma$ with $|\gamma|=\lfloor\frac{n+q-1}{2}\rfloor -n+1$. A quick proof comes from the following generating function
$$\frac{x^l}{(1-x)^{l+1}}=\sum\_{p=0}^{\infty} \binom{p}{l}x^p$$ and looking at the coefficient of $x^{n+q-1}$ in the identity
$$x^{n-1}\prod\_{i=1}^n \frac{x^{\gamma\_i}}{(1-x)^{1+\gamma\_i}}=\frac{x^{|\gamma|+n-1}}{(1-x)^{n+|\gamma|}}$$


