summation over partial multinomial coefficients So the standard multinomial theorem states that
$$
\sum_{i_1,\ldots,i_m} \binom{n}{i_1,\ldots,i_m}\cdot [i_1+\cdots +i_m=n]=m^n
$$
where the summation is over non-negative numbers and $[\cdot]$  is the indicator function that equals $1$ if and only if $\cdot$ is satisfied and $0$ otherwise. Now, do we know anything about say (for even $m$ and $n$ being a power of $2$),
$$
\sum_{i_1,\ldots,i_m} \binom{n}{i_1,\ldots,i_m}\cdot [i_1+\cdots +i_{m/2+1}\geq n/2]\cdot [i_1+\cdots +i_m=n]?
$$
or
$$
\sum_{i_1,\ldots,i_m} \binom{n}{i_1,\ldots,i_m}\cdot [i_1+\cdots +i_{m/2+2}\geq 3n/4]\cdot [i_1+\cdots +i_m=n]?
$$
or in general
$$
\sum_{i_1,\ldots,i_m} \binom{n}{i_1,\ldots,i_m}\cdot [i_1+\cdots +i_{m/2+k}\geq (2^k-1)n/2^k]\cdot [i_1+\cdots +i_m=n]?
$$
for arbitrary $k\in \{1,\ldots,m/2\}$? Eventually for large enough $k$ this just converges to the standard multonimial theorem statement.
In general, I'm looking to understand or upper bound the following quantity
$$
\sum_{i_1,\ldots,i_m} \binom{n}{i_1,\ldots,i_m}\cdot \prod_{k=1}^{m/2}[i_1+\cdots +i_{m/2+k}\geq (2^k-1)n/2^k]
$$
Any pointers/help would be appreciated!
 A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
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$\ds{m\ \mbox{and}\ n}$ are even numbers. Then, lets set
$m \equiv 2M$ and $n = 2N$ where $\ds{M, N, \in \mathbb{N}_{\ \geq\ 0}}$.
Lets study one of the above expressions. Namely, the first non standard one:
\begin{align}
&\bbox[5px,#ffd]{\sum_{k_{1},\ldots,k_{2m}}{2N \choose k_{1},\ldots,k_{2M}}
\bracks{k_{1} + \cdots + k_{M + 1} \geq N}}
\\[5mm] = &
\sum_{s = N}^{\infty}\sum_{k_{1},\ldots,k_{2M}\ =\ 0}^{\infty}{\pars{2N}! \over k_{1}!,\ldots,k_{2M}!}
\bracks{k_{1} + \cdots + k_{M + 1} = s}
\\[5mm] = &\
\pars{2N}!\sum_{s = N}^{\infty}\sum_{k_{1},\ldots,k_{2M}\ =\ 0}^{\infty}{1 \over k_{1}!,\ldots,k_{2M}!}
\bracks{z^{\large s}}z^{k_{1} + \cdots + k_{M + 1}}
\\[5mm] = &\
\pars{2N}!\sum_{s = N}^{\infty}\bracks{z^{\large s}}
\pars{\sum_{k = 0}^{\infty}{z^{k} \over k!}}^{M + 1}
\pars{\sum_{q = 0}^{\infty}{1 \over q!}}^{M - 1}
\\[5mm] = &\
\pars{2N}!\expo{M - 1}\sum_{s = N}^{\infty}\bracks{z^{\large s}}
\expo{\pars{M + 1}z}
\\[5mm] = &\
\bbx{\pars{2N}!\,\expo{M - 1}
\sum_{s = N}^{\infty}{\pars{M + 1}^{s} \over s!}} \\ &
\end{align}
