Stirling Numbers of second kind recurrence

Question

While learning the stirling numbers of second kind, we see a recursive relation : $$S(n, r) = S(n-1, r-1) + k\cdot S(n-1, r)$$

And we get its solution as : $$S(n, r) = (1/{r!})[r^n - C(r,1)\cdot (r-1)^n + \cdots + (-1)^{r-1}\cdot C(r,r-1)\cdot 1^n]$$

Can you please tell me how the above solution of the recurrence written is obtained? Thanks in advance.

For the term inside the brackets, use the Inclusion-Exclusion Principle to count the number of ways of distributing $n$ objects in $r$ labeled boxes so that no box left empty. The factor of $1/r!$ is used to account for the fact that the boxes are not labeled. — N. F. Taussig, Mar 21 '21 at 10:13
The inclusion-exclusion calculation used to derive the closed form can be found in this answer. — Brian M. Scott, Mar 21 '21 at 23:21

score 2 · Accepted Answer · answered Mar 21 '21 at 17:52

A set partition into $k$ parts is unordered partition of a set $S$ in unordered collection pairwise disjoint sets $S_1,S_2,...,S_k$. Number of set partitions of ${1,2,...,n}$ into $k$ parts is $S(n,k)$. Also a set partition of ${1,2,...,n}$ is equivalent to $ \cdot \times Seq\{1\} \times \cdot \times seq\{1,2\} \times \cdot \times \ldots \times seq\{1,2,\ldots,k\}$ where $\cdot$ and all numbers $j$ have size $1$. To see this if a number $l$ is at position $j$ then you put $j$ in $S_l$ and if $i^{th}$ dot is at position $j$ then you put $j$ in $S_i$. So $$ \sum_{n\geq 0 } S{(n,r)}x^n = \dfrac{x}{1-x}\times\dfrac{x}{1-2x}\times\ldots\times\dfrac{x}{1-rx}$$ Extracting coefficient of $x^n$ you get your formula.

Stirling Numbers of second kind recurrence

1 Answers1