> On Mar 29, 2018, at 6:48 PM, Ranjan Maitra <mai...@email.com> wrote:
>
> Dear friends,
>
> I would like to get all possible arrangements of n objects listed 1:n on a
> circle.
>
> Now this is easy to do in R. Keep the last spot fixed at n and fill in the
> rest using permuations(n-1, n-1) from the gtools package.
>
> However, what if clockwise or counterclockwise arrangements are the same? I
> know that half of the above (n - 1)! arrangements are redundant.
>
> Is there an easy way to list these (n-1)!/2 arrangements?
>
Well half of these arrangements will be of the form
`k, ... , j, n'
and half will be of the form
`j, ..., k, n'
So fix n in position n, select (k,j), and require that the first position is
min(k,j) and position n-1 is max(k,j).
There are choose(n-1,2) choices for {(k,j):k<j!=n}. Then you have (n-3)! ways
to fill the rest.
That gives (n-1)!/((n-3)! * 2!) * (n-3)! = (n-1)!/2 arrangements.
HTH,
Chuck
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