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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.

urpi
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1 Answers1

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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 $k!$.

Brian M. Scott
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