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According to page

If clock-wise and anti-clock-wise orders are taken as not different, then total number of circular-permutations is given by (n-1)!/2!

But how it can be correct? For example when n=1 answer should be 1. But formula will give .5

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for $n\geq2$ only, the question of clockwise and anti clockwise permutation comes into picture. – Abhra Abir Kundu Feb 3 '13 at 9:01

1 Answer 1

up vote 1 down vote accepted

You're right; that page fails to mention that the result is only valid for $n\ge3$. For $n\lt3$, the argument that each permutation is counted twice in $(n-1)!$ isn't valid, since an arrangement of $1$ object is transformed into itself by reflection and an arrangement of $2$ objects is transformed into a cyclically equivalent arrangement by reflection, and cyclic equivalence is already taken into account in $(n-1)!$.

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