I have been stuck with the following problem for hours, and I was hoping someone could give me a hint to attack it. Thanks! We define a distribution as a function together with a linear ordering on the preimage of each element of the codomain. Using exponential generating functions, we want to obtain the formula $$ {n \choose k} (n-1)_{n-k}$$ for the number of surjective distributions from a set n labelled objects to a set of k labelled places.
Particularly, I am trying to solve the problem by using the theory of species. I know that the number of surjections from a $n$-element set to a $k$-element set is $k!S(n,k)$, where $S(n,k)$ is the Stirling number of the second kind, but I don't know how to combine this with the fact that we have the linear ordering on the preimage of each element of the codomian.
That formula appears to be incorrect. There are $12$ distributions from $[3]$ to $[2]$, not
$$\binom32(3-1)^{\underline{3-2}}=6\;.$$
There is a bijection between distributions from $[n]$ to $[k]$ and ordered pairs $\langle\pi,K\rangle$ such that $\pi$ is a permutation of $[n]$ and $K$ is a $(k-1)$-subset of the $n-1$ spaces between adjacent elements of $\pi$. For example, the distribution from $[6]$ to $[3]$ that sends $1,3$, and $4$ to $1$ and orders them $314$, sends $6$ to $2$, and sends $2$ and $5$ to $3$ and orders them $52$ corresponds to $314|6|52$, where $\pi=314652$ and the bars show which $2$ spaces are in $K$. There are $n!$ permutations and $\binom{n-1}{k-1}$ ways to choose $K$, so there are
$$\binom{n-1}{k-1}n!=\binom{n-1}{n-k}n!=\frac{(n-1)^{\underline{n-k}}}{(n-k)!}n!=k!\binom{n}k(n-1)^{\underline{n-k}}$$
distributions from $[n]$ to $[k]$. Thus, your formula is missing a factor of $k!$.