Hi Rupesh,
Despite the fact that the answer may not help you, my colleagues and I
worked on this problem today. The closed-form solution for two lists
of size n is as follows:
The number of possible permutations, following your original
instructions, are:
P(n) = (2*n)! / (n!)^2 - (2*n)! / [(n-2)!*(n+2)!] = (2*n)nCr(n) -
(2*n)nCr(n-2)
In practice, this number gets very large very quickly. For example,
the result for n={1,2,...14} is
P(n) = {1, 2, 5, 14, 42, 132, 429, 1430, 4862, 16796, 58786, 208012,
742900, 2674440}
The reasons for this are complicated, but contact me if you're
interested and I'd be glad to elaborate.
Mike
On Jul 17, 11:04 pm, "Rupesh Bhochhi" <[EMAIL PROTECTED]> wrote:
> Thank you Karthik for your effort. I figure out lately that My
> question is not that much logical. Actually there will be only one
> possible combination that we can easily get as we have already got the
> sorted lists.
>
> On 7/17/07, Karthik Singaram L <[EMAIL PROTECTED]> wrote:
>
>
>
> > Hi,
> > Although, I cannot give you a complete solution right now..I will give you
> > the following relations (I am not sure If they are correct.....discussions
> > welcome)
>
> > T(0, x) = 1
> > T(x, 0) = 1
> > T(1, x) = (x+1)
> > T(x, 1) = (x+1)
> > T(n, m) = Sum[i=0..n] { T(n-i, m-2) * (i+1) }
>
> > If you are looking for a program, then you can code this up...If you are
> > looking for a closed form solution, this needs to be tidied up.
>
> --
> Rupesh Bhochhibhoya, [EMAIL PROTECTED]
> Oklahoma State University
> Stillwater, OK 74075
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