Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

This question is prefaced by this part:

We use the notation S(k,n) to stand for the number of partitions of a k element set with n blocks. For historical reasons, S(k,n) is called a Stirling Number of the second kind.


In a partition of the set [k], the number k is either in a block by itself, or it is not. How does the number of partitions of [k] with n parts in which k is in a block with other elements of [k] compare to the number of partitions of [k - 1] into n blocks? Find a two-variable recurrence for S(k,n), valid for k and n larger than one.

So if I understand this correctly the answer will be a proof? Can I reach the answer by enumerating different example partitions of sets? I appreciate any tips and help. Thank you

share|cite|improve this question
up vote 2 down vote accepted

The answer will be a recurrence together with an explanation justifying that recurrence. For example, the binomial coefficients satisfy the two-variable recurrence $$\binom{k}n=\binom{k-1}{n-1}+\binom{k-1}n\tag{1}$$ (which you may recognize as the rule used to form Pascal’s triangle). The Stirling numbers of the second kind, which are also written $$\left\{\matrix{k\\n}\right\}$$ instead of $S(k,n)$, satisfy a two variable recurrence quite similar to $(1)$, though a little more complicated. One term counts the partitions of $[k]$ in which $\{k\}$ is a block by itself; the other counts the partitions of $[k]$ in which $k$ is in a block with something else.

To get you started, if $\{k\}$ is a block by itself, the remaining $k-1$ elements of $[k]$ must make up the other $n-1$ blocks. There are $S(k-1,n-1)$ ways in which they can do this, so there are how many partitions of $[k]$ in which $\{k\}$ is a block by itself?

share|cite|improve this answer
Binet's formula, isn't that about Fibonacci numbers? I know this one under the name of Pascal's rule. – Marc van Leeuwen Feb 22 '12 at 8:17
@Marc: My brain evidently went walkabout for a moment. Thanks! – Brian M. Scott Feb 22 '12 at 8:19
@Brian, Thank You Very Much! I'm slowly getting it, but still a bit puzzled(just the concept of Stirling #s itself), so I'll have to read more on wikipedia. – Adel Feb 22 '12 at 21:59

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.