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How many ways we can divide '$n$' distinct items into $r$ identical groups where a group must have at-least one item?

I am looking for some approach for solving this problem.

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Could you please specify what you mean by "identical"? – Rasmus Aug 21 '11 at 18:53
@Rasmus:"identical" = "indistinguishable". – Quixotic Nov 7 '11 at 1:00
So, if you say "identical group" you mean "group consisting of identical elements"? – Rasmus Nov 7 '11 at 9:23
up vote 3 down vote accepted

Your question is addressed exactly via the Stirling Numbers of the Second Kind. In particular, there is an explicit formula $$\left\{\begin{matrix} n \\ r \end{matrix}\right\} = \frac{1}{r!}\sum_{j=0}^{r}(-1)^j{r \choose j} (r-j)^n.$$

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But the OP wants the subsets to be identical (whatever he means by that). – Rasmus Aug 21 '11 at 18:44
I took "identical" to mean "indistinguishable" or "unlabeled". – Austin Mohr Aug 21 '11 at 18:48
and if the groups are also distinct then we have to just multiply this with $r!$ which is $r! \times \left\{\begin{matrix} n \\ r \end{matrix}\right\}$ – Quixotic Nov 7 '11 at 1:05

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