Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
1  
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
add comment

1 Answer

up vote 2 down vote accepted

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

share|improve this answer
    
But the OP wants the subsets to be identical (whatever he means by that). –  Rasmus Aug 21 '11 at 18:44
1  
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
add comment

Your Answer

 
discard

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.