Difference between revisions of "Stirling number"
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For example, <math>S_2(4, 2) = 7</math> because we have the partitions <math>\{1 | 234, 2|134, 3|124, 4|123, 12|34, 13|24, 14|23\}</math>. | For example, <math>S_2(4, 2) = 7</math> because we have the partitions <math>\{1 | 234, 2|134, 3|124, 4|123, 12|34, 13|24, 14|23\}</math>. | ||
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| + | They can be calculated using the following formula: | ||
| + | <cmath> S(n,k) = \frac{1}{k!}\sum_{j=0}^{k}(-1)^j{k \choose j} (k-j)^n </cmath> | ||
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[[Category:Combinatorics]] | [[Category:Combinatorics]] | ||
Revision as of 11:13, 25 October 2010
There are two kinds of Stirling numbers: Stirling numbers of the first kind and Stirling numbers of the second kind. They appear in many situations in combinatorics.
Stirling Numbers of the First Kind
The Stirling number of the first kind, 
, is the number of permutations of an 
-element set with exactly 
cycles.
For example, 
because (writing all our permutations in cycle notation) we have the permutations 
.
Stirling Numbers of the Second Kind
The Stirling number of the second kind, 
, is the number of partitions of an -element set into exactly subsets.
For example, 
because we have the partitions 
.
They can be calculated using the following formula:
![\[S(n,k) = \frac{1}{k!}\sum_{j=0}^{k}(-1)^j{k \choose j} (k-j)^n\]](http://latex.artofproblemsolving.com/8/1/6/816df54552ad7f936caa67b6dfb49c4613434ed7.png)
