Simplify $\sum_{k = 0}^n\sum_{k_1+\ldots+k_{m}=n-k} \binom{nm}{k_1,\ldots,k_m,n-k_1,\ldots,n-k_m}$

Question

Let $n$ and $m$ be positive integers. I want to find a formula for the following expression:

$$\sum_{k = 0}^n\quad \sum_{k_1+...+k_m=n-k} \quad \binom{n-k}{k_1,\ldots,k_m}\binom{nm}{n-k_1,\ldots,n-k_m,n-k}$$

$$= \sum_{k = 0}^n \quad \sum_{k_1+...+k_{m}=n-k} \quad \binom{nm}{k_1,\ldots,k_m,n-k_1,\ldots,n-k_m}$$

I am wondering if there are any special identities I could use to simplify this expression.

Answer 1

This is not an answer. Use

$$\sum_{k_0=0}^n\sum_{k_1+\cdots+k_m=n-k_0}\leftrightarrow \sum_{k_0+k_1+\cdots + k_m = n}$$ which is notationally seemingly odd, yet in the first $k_0$ is fixed and in the second $k_0$ varies. This second one seems useful:

$$(x_0+\cdots+x_m)^n=\sum_{k_0+\cdots+k_m=n}\binom{n}{k_0,k_1,...,k_m}x^{k_0}_0x_1^{k_1}\cdots x_m^{k_m}$$