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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 Steven Gerhard in advance!

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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 $\binom{\alpha}{\gamma}=\frac{\alpha !}{\gamma! (\alpha-\gamma)!}$ where $\alpha!=\alpha_1 !\alpha_2! \cdots \alpha-\gamma)!}=\prod_{i=1}^n \alpha_n!$.

The answer should be 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$. Could anyone help? (I don't some work with Here is the solution for$n=1$. The sum$\sum_{|\alpha|=q}\binom{\alpha}{\gamma}$is the single term$\binom{q}{\gamma}$, so the monotonicity of the binomial coefficients, but I don't know how distribution gives the maximum at$\binom{q}{\lfloor q/2\rfloor}$. For$n=2$, the problem amounts to generalize my method maximizing$\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}$. More rewrites and updates to multinomials.)come. Thanks to Steven in advance! 2 added 159 characters in body I'm trying to find the following maximum:$\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)!}$where$\alpha!=\alpha_1 !\alpha_2! \cdots \alpha_n!$. The answer should be in terms of$n$and$q\$. Could anyone help? (I don't some work with binomial coefficients, but I don't know how to generalize my method to multinomials.)

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