A can remain same in following cases:
-> If m, n are equal in all the N iterations
-> if m, n are equal in N-2 iterations but in 1 iteration m and n both are
different in that case there should be one iteration where m and n should be
same as the previous
iteration so that they are swapped again to the same position.
In the similar fashion if you think, the pairwise similar iteration should
be there.
Now, if in all N iterations probability that m = n is: (N/N*N) *
(1/N)*...(1/N) = (1/N)^N
two of the iterations have different m and n then probability is: NC2 *
(1/N)^(N-2) * {(1 - 1/N)* 1/NC2} = NC2 * (1/N)^N * 2
similarly, four of the iterations have different m and n then probability
NC4 * (1/N)^(N-4) *{(1 - 1/N) * 1/NC2}^2 = NC4 * (1/N)^N * 2^2
-Piyush
On Mon, Sep 12, 2011 at 10:24 PM, payal gupta <[email protected]> wrote:
> @piyush..
> cud u plzz..xplain...n elaborate
>
> Regards,
>
> Payal Gupta
>
>
> On Mon, Sep 12, 2011 at 10:15 PM, Piyush Grover <[email protected]
> > 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 <
>> [email protected]> 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 <[email protected]
>>> > 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 <[email protected]>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 <[email protected]
>>>>> > 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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