Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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

share|cite|improve this question
for $n\geq2$ only, the question of clockwise and anti clockwise permutation comes into picture. – Abhra Abir Kundu Feb 3 '13 at 9:01
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 $(n-1)!$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.