Question: Given $x\_0 \ldots x\_k$ and $n$, Define $$f(Q)=\sum\_{\substack{n\_0+\ldots+n\_k=n \\ n\_0,\ldots,n\_k \ >=0 \\n\_1+2\*n\_2+\ldots+k\*n\_k=Q}} \binom{n}{n\_0,\cdots,n\_k}x\_{0}^{n\_0}\ldots x\_{k}^{n\_k}$$. Note that $$\sum\_{Q=0}^{n\*k} f(Q)=(\sum\_{i=0}^{k}x\_i)^n$$ which comes from the multinomial expansion. I was wondering how to calculate $\sum\_{Q=0}^{n\*k} Q\cdot f(Q)$.

I've checked several small cases, and it seems that $\sum\_{Q=0}^{n\*k} Q\cdot f(Q) = n(\sum\_{i=0}^{k}x\_i)^{n-1}(\sum\_{i=0}^{k}ix\_i)$, but I'm not sure how to prove that.

Answer: 
Wikipedia on [Bell polynomials](https://en.wikipedia.org/wiki/Bell_polynomials):

> 
> Likewise, the partial ordinary Bell polynomial, in contrast to the usual exponential Bell polynomial defined above, is given by $$\hat{B}\_{n,k}(x\_1,x\_2,\ldots,x\_{n-k+1})=\sum \frac{k!}{j\_1! j\_2! \cdots j\_{n-k+1}!} x\_1{}^{j\_1} x\_2{}^{j\_2} \cdots x\_{n-k+1}{}^{j\_{n-k+1}},$$
>  where the sum runs over all sequences $j\_1, j\_2, j\_3, \ldots, j\_{n−k+1}$ of non-negative integers such that $$
> j\_1 + j\_2 + \cdots + j\_{n-k+1} = k, \\
> j\_1 + 2j\_2 + \cdots + (n-k+1)j\_{n-k+1} = n.$$
> 
> 
> 

So your sum $$\sum\_{\substack{n\_0+\ldots+n\_k=n \\ n\_0,\ldots,n\_k \ >=0 \\n\_1+2n\_2+\ldots+kn\_k=Q}} \binom{n}{n\_0,\cdots,n\_k}x\_{0}^{n\_0}\ldots x\_{k}^{n\_k}$$
is
$$\sum\_{n\_0=0}^n \binom{n}{n\_0} x\_{0}^{n\_0} \hat{B}\_{Q,n-n\_0}(x\_1, x\_2, \ldots, x\_k)$$

---

Using $${\hat {B}}\_{n,k}(x\_{1},x\_{2},\ldots ,x\_{n-k+1})={\frac {k!}{n!}}B\_{n,k}(1!\cdot x\_{1},2!\cdot x\_{2},\ldots ,(n-k+1)!\cdot x\_{n-k+1})$$ we have

$$\frac{n!}{Q!}\sum\_{n\_0=0}^n \frac{x\_{0}^{n\_0}}{n\_0!} B\_{Q,n-n\_0}(1!\cdot x\_1, 2!\cdot x\_2, \ldots, k!\cdot x\_k)$$

Then the recurrence relation $$B\_{n,k}=\sum\_{i=1}^{n-k+1} \binom{n-1}{i-1} x\_i B\_{n-i,k-1}$$

gives you an evaluation strategy.
Starting from Hamed's answer, we can give a quick proof that $\sum\_{Q\ge 0} Qf(Q)$ equals what you think it is. They proved
$$
\sum\_{n\ge 0}\sum\_{q\ge 0}\frac1{n!}f\_n(Q)X^nY^q = \exp\Big(\sum\_{m=0}^k x\_mXY^m\Big)
$$
Differentiating both sides with respect to $Y$, and then multiplying by $Y$, you get
$$
\sum\_{n\ge 0}\sum\_{q\ge 0}\frac1{n!}Qf\_n(Q)X^nY^q = \big(\sum\_{k=0}^m mx\_m XY^m\big)\exp\Big(\sum\_{m=0}^k x\_mXY^m\Big)
$$
Notice that we now have a $Qf\_n(Q)$. These need to be summed, so set $Y=1$:
$$
\sum\_{n\ge 0}\sum\_{q\ge 0}\frac1{n!}Qf\_n(Q)X^n = \big(\sum\_{k=0}^m mx\_m X\big)\exp\Big(X\sum\_{m=0}^k x\_m\Big)
$$
Finally, extract the coefficient of $X^n$ of both sides. On the left hand side, $X\_n$ has a $\frac1{n!}\sum\_{q\ge 0}QF\_n(Q)$ attached to it. On the right, you get the $X^n$ coefficient by multiplying $\big(\sum\_{k=0}^m mx\_m\big)$ with the $X^{n-1}$ coefficient of $\exp\Big(X\sum\_{m=0}^k x\_m\Big)$. This implies
$$
\frac1{n!}\sum\_{q\ge 0}QF\_n(Q)=\big(\sum\_{k=0}^m mx\_m\big)\cdot \frac1{(n-1)!}\Big(\sum\_{k=0}^m x\_m\Big)^{n-1}
$$
which is what you wanted.