Explicit formulas for Stirling numbers of second kind

Question

I have seen some recursive formulas for Stirling I and II kind but I am especially interested in explicit formulas.

I was thinking and get: $$ \left\{ {m \atop n} \right\} = \sum_{k_i>0 \wedge k_1+k_2+...+k_n = m} \frac{m!}{n!}\cdot \frac{1}{k_1! \cdot k_2! \cdot ... \cdot k_n!}$$ I am not sure if it is ok but supposedly it is (I created it with combinatorics interpretation)
Can somebody look at that and tell me if it is ok/wrong?

By the way,if you have some nice formulas, I will be grateful for sharing that.

Answer 1

The proposed formula is correct: see on wikipedia the exponential generating function for the Stirling numbers of second kind: \begin{align*} \sum_{n\ge k} {n \brace k}\frac{x^n}{n!}&= \frac{\left(e^x-1\right)^k}{k!}\\ &=\frac{1}{k!}\left(\sum_{j\ge1}\frac{x^j}{j!}\right)^k.\end{align*} Just expand the right hand side and equate the coefficient of $x^n$ on both sides.

Answer 2

A probabilistic proof/interpretation (perhaps this is how you got it?). We consider the experiment of placing $m$ balls into $n$ urns (with uniform probability). The distribution is the multinomial: $$P(x_1,x_2 \cdots x_n) = \frac{m!}{x_1 ! x_2! \cdots x_n!} \frac{1}{n^m} \, \left[\sum{x_i}=m\right] \left[0\le x_i \le m \right] \tag1$$

Now, consider the event $E$ that all urns are not empty. Its probability, by the combinatorial interpretation of Stirling numbers of the second kind, is $$P(E)=P(x_1>0,x_2>0\cdots)= \frac{n!{m \brace n}}{n^m} \tag2$$

But from $(1)$

$$P(E)= \sum_{x_i>0 \wedge x_1+x_2+...+x_n = m} \frac{m!}{n^m}\frac{1}{x_1! \cdot x_2! \cdot ... \cdot x_n!} \tag3$$

Equating the RHS of $(2)$ and $(3)$ we get your formula.