0
$\begingroup$

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.

  • 1
    Could you please specify what you mean by "identical"?2011-08-21
  • 0
    @Rasmus:"identical" = "indistinguishable".2011-11-07
  • 0
    So, if you say "identical group" you mean "group consisting of identical elements"?2011-11-07

1 Answers 1

3

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

  • 0
    But the OP wants the subsets to be identical (whatever he means by that).2011-08-21
  • 1
    I took "identical" to mean "indistinguishable" or "unlabeled".2011-08-21
  • 0
    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\}$2011-11-07