@Puneet, you are right but we can have only n-1 dividers.

On Wed, Aug 17, 2011 at 4:15 PM, Puneet Goyal <puneetgoya...@gmail.com>wrote:

> I think it should be 2^n -1
>
> Explanation
> We can visualize it as n balls are placed and we have to place some
> dividers (max=n) in betweek to divide them into groups.
>
> If we choose no divider its nC0 , but we dont have to include it
> With 1 divider its nC1
> and so on..
> So the total no. of ways will be
> (nC0+nC1+nC2..........nCn)-nC0= 2^n-1
>
> Regards,
> Puneet
>
> On Wed, Aug 17, 2011 at 4:05 PM, Rohit Srivastava 
> <access2ro...@gmail.com>wrote:
>
>> +1 to nitin
>>
>>
>> On Wed, Aug 17, 2011 at 2:48 PM, Vijay Kansal <vijaykans...@gmail.com>wrote:
>>
>>> my bad 2^(n-1)...
>>>
>>> On Aug 17, 2:17 pm, Vijay Kansal <vijaykans...@gmail.com> wrote:
>>> > @nitin it must be 2^n i think
>>> >
>>> > On Aug 17, 3:48 am, Bharat Kul Ratan <bharat.kra...@gmail.com> wrote:
>>> >
>>> >
>>> >
>>> >
>>> >
>>> >
>>> >
>>> > > It might be useful:
>>> http://www.artofproblemsolving.com/Wiki/index.php/Partition_%28combin...
>>>
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>
>
>
> --
> -----------------------------------------------------------------------
> Puneet Goyal
> Student of B. Tech. III Year (Software Engineering)
> Delhi Technological University, Delhi
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