@piyush..
cud u plzz..xplain...n elaborate
Regards,
Payal Gupta
On Mon, Sep 12, 2011 at 10:15 PM, Piyush Grover
<piyush4u.iit...@gmail.com>wrote:
> Sry i was little wrong:
>
> nC0*(1/n)^n + nC2 *2*(1/n)^n + nC4*2^2*(1/n)^n+....+nCn*2^(n/2)*(1/n)^n
> when n is even
>
> nC0*(1/n)^n + nC2 *2*(1/n)^n +
> nC4*2^2*(1/n)^n+....+nCn-1*2^(n-1/2)*(1/n)^n when n is odd
>
>
> On Mon, Sep 12, 2011 at 10:08 PM, Piyush Grover <piyush4u.iit...@gmail.com
> > wrote:
>
>> it should be:
>>
>> (1/n)^n * (1 + 2 + 2^2 + 2^3 +....(n/2)+1 terms) = {2^(1 + n/2) -
>> 1}*(1/n)^n when n even
>>
>> (1/n)^n * (1 + 2 + 2^2 + 2^3 +....(n-1/2)+1 terms) = {2^(1 + (n-1)/2) -
>> 1}*(1/n)^n when n is odd
>>
>>
>> -Piyush
>>
>>
>> On Mon, Sep 12, 2011 at 8:01 PM, sandeep nagamalli
>> <nsandee...@gmail.com>wrote:
>>
>>> I think it is 1 / (2N)
>>>
>>> (1/N) * (1/N)*(N/2) = 1/(2N)
>>>
>>>
>>> On Mon, Sep 12, 2011 at 6:33 PM, Akash Mukherjee <akash...@gmail.com>wrote:
>>>
>>>> this is a case, but isnt der a case of circular permutation 2??
>>>> what say abt dis
>>>> 0,1
>>>> 2,3
>>>> 4,5
>>>> 0,1
>>>> 2,3
>>>> 4,5
>>>>
>>>> it wrks i gues??
>>>>
>>>>
>>>> On Mon, Sep 12, 2011 at 12:56 PM, teja bala
>>>> <pawanjalsa.t...@gmail.com>wrote:
>>>>
>>>>>
>>>>>
>>>>> I think it is 1/N
>>>>>
>>>>> let N=6 that means rand(6)= takes values 0-5 i.e 6 values.
>>>>> rand(m)=6 values
>>>>> rand(n)=6 values
>>>>> total combinations 6*6=36 values set but among dem array will change
>>>>> only for
>>>>> (0,0)(1,1)(2,2)(3,3)(4,4)(5,5)(6,6)=6values of N
>>>>>
>>>>> So 6/36=1/6
>>>>>
>>>>> If we generalize it 1/N
>>>>>
>>>>> correct me if i'm wrong?
>>>>>
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>>>>
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>>>
>>>
>>>
>>> --
>>> Thanks&Regards:
>>> -sandeep
>>>
>>>
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>>
>>
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