@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? >>>>> >>>>> -- >>>>> You received this message because you are subscribed to the Google >>>>> Groups "Algorithm Geeks" group. >>>>> To post to this group, send email to [email protected]. >>>>> To unsubscribe from this group, send email to >>>>> [email protected]. >>>>> For more options, visit this group at >>>>> http://groups.google.com/group/algogeeks?hl=en. >>>>> >>>> >>>> -- >>>> You received this message because you are subscribed to the Google >>>> Groups "Algorithm Geeks" group. >>>> To post to this group, send email to [email protected]. >>>> To unsubscribe from this group, send email to >>>> [email protected]. >>>> For more options, visit this group at >>>> http://groups.google.com/group/algogeeks?hl=en. >>>> >>> >>> >>> >>> -- >>> Thanks&Regards: >>> -sandeep >>> >>> >>> -- >>> You received this message because you are subscribed to the Google Groups >>> "Algorithm Geeks" group. >>> To post to this group, send email to [email protected]. >>> To unsubscribe from this group, send email to >>> [email protected]. >>> For more options, visit this group at >>> http://groups.google.com/group/algogeeks?hl=en. >>> >> >> > -- > You received this message because you are subscribed to the Google Groups > "Algorithm Geeks" group. > To post to this group, send email to [email protected]. > To unsubscribe from this group, send email to > [email protected]. > For more options, visit this group at > http://groups.google.com/group/algogeeks?hl=en. > -- You received this message because you are subscribed to the Google Groups "Algorithm Geeks" group. To post to this group, send email to [email protected]. To unsubscribe from this group, send email to [email protected]. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
