# Stirling numbers

2010 Mathematics Subject Classification: *Primary:* 05A *Secondary:* 11B73 [MSN][ZBL]

In combinatorics, counts of certain arrangements of objects into a given number of structures. There are two kinds of Stirling number, depending on the nature of the structure being counted.

The *Stirling numbers of the first kind* $S(n,k)$ count the number of ways $n$ labelled objects can be arranged into $k$ cycles: cycles are regarded as equivalent, and counted only once, if they differ by a cyclic permutation, thus $[ABC] = [BCA] = [CAB]$ but is counted as different from $[CBA] = [BAC] = [ACB]$. The order of the cycles in the list is irrelevant.

For example, 4 objects can be arranged into 2 cycles in eleven ways, so $S(4,2) = 11$.

- [ABC],[D]
- [ACB],[D]
- [ABD],[C]
- [ADB],[C]
- [ACD],[B]
- [ADC],[B]
- [BCD],[A]
- [BDC],[A]
- [AB],[CD]
- [AC],[BD]
- [AD],[BC]

The number of permutations on $n$ letters with exactly $k$ cycles is given by $S(n,k)$. Hence $$ n!=\sum_{k=1}^nS(n,k) \ . $$

The *Stirling numbers of the second kind* $s(n,k)$ count the number of ways $n$ labelled objects can be arranged into $k$ subsets: the subsets are regarded as equivalent, and counted only once, if they have the same elements, thus $\{ABC\} = \{BCA\} = \{CAB\} = \{CBA\} = \{BAC\} = \{ACB\}$; the order of the subsets in the list is also irrelevant.

For example, 4 objects can be arranged into 2 subsets in seven ways, so $s(4,2) = 7$:

- {ABC},{D}
- {ABD},{C}
- {ACD},{B}
- {BCD},{A}
- {AB},{CD}
- {AC},{BD}
- {AD},{BC}

If $B_n$ denotes the Bell numbers, the total number of partititons of a set of size $n$, thren $$ B_n = \sum_{k=1}^n s(n,k) \ . $$

See also: Combinatorial analysis.

## References

- Ronald L. Graham, Donald E. Knuth, Oren Patashnik,
*Concrete Mathematics*, Addison Wesley (1989) pp.243-253. ISBN 0-201-14236-8

**How to Cite This Entry:**

Stirling numbers.

*Encyclopedia of Mathematics.*URL: http://encyclopediaofmath.org/index.php?title=Stirling_numbers&oldid=40063