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Let $m$ and $n$ be two non-negative integers. I am interested in finding a closed formula or an asymptotic approximation of the following sum $$ S_{n,m} = \sum_{k_1+\dots+k_m = n}\max(k_1,\dots,k_m)\binom{n}{k_1,\dots ,k_m} $$ Any hint would be much appreciated. Thanks.

A good asymptotic approximation with respect to $n$ (respectively $m$) while $m$ (respectively $n$) is fixed is also fine.

Edit: I found out that this question has been asked on mathoverflow more than 3 years ago but has only been answered to partially. This may mean that there is a chance a solution is already lurking somewhere or the problem itself might be quite hard.

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A partial answer with a simple lower bound: $$\begin{eqnarray*}\sum_{k_1+\ldots+k_m=n}\max(k_1,\ldots,k_m)\binom{n}{k_1,\ldots,k_m}&\color{red}{\geq}& \frac{1}{m}\sum_{k_1+\ldots+k_m=n}(k_1+\ldots+k_m)\binom{n}{k_1,\ldots,k_m}\\ &=&\frac{1}{m}\,{\large\nabla\cdot}\!\!\!\left.\sum_{k_1+\ldots+k_m=n}x_1^{k_1}\cdots x_m^{k_m}\binom{n}{k_1,\ldots,k_m}\right|_{\bar{x}=1}\\&=&\frac{1}{m}\,\left.{\nabla\cdot}(x_1+\ldots+x_m)^n\right|_{\bar{x}=1}\\&=&\color{red}{n\,m^{n-1}}.\end{eqnarray*}$$ Here $\nabla\cdot$ is the divergence operator.

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    Thanks Jack but I already know this lower bound and the generous upper bound obtained from $\max(k_1,\dots,k_m)\leq n$. I'm really interested in knowing the asymptotic behaviour more precisely. – 2017-01-18
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    @Alex: I think that more accurate approximations comes from replacing the binomial/multinomial distribution with a normal distribution. It is reasonable to expect that the greatest contribute comes from the $m$-uples $(k_1,\ldots,k_m)$ for which every $k_i$ is reasonably close to $\frac{n}{m}$. – 2017-01-18
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    On the other hand, the generous upper bound is $n m^{n}$, that is just $m$ times the lower bound computed from the fact that the maximum is $\geq$ than the average. – 2017-01-18
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    So, for sure, the asymptotic behaviour is given by $$ n m^{m-1+\lambda(m,n)}$$ where $0\leq\lambda(m,n)\leq 1$. – 2017-01-18
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    I assume the power is $n-1 + \lambda(m,n)$ of course and it is even safe to assume the bounds for $\lambda$ are strict. Now, how does that $\lambda$ behave asymptotically? does it decrease to $0$ if we fix $m$ and increase $n$? – 2017-01-18
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    @Alex: my first bet would be that $\lambda$ is close to $\frac{1}{2}$, since I am expecting that a $\sqrt{m}$ factor arises from the computation of the standard error of the involved multinomial distribution. However, I might be wrong. – 2017-01-18
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    So to get a better approximation of that $\lambda$, can you elaborate a bit on how to the approximation with the normal distribution you mentioned above can be used? – 2017-01-18
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    @Alex: I was referring to the multinomial analogue of this well-known fact: http://www.real-statistics.com/binomial-and-related-distributions/relationship-binomial-and-normal-distributions/ – 2017-01-18
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    Let us [continue this discussion in chat](http://chat.stackexchange.com/rooms/51972/discussion-between-alex-and-jack-daurizio). – 2017-01-18