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Hey please help me with this question... Find the number of circular permutation of the word 'CIRCULAR'. Number of circular permutaion is (n-1)!

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closed as off-topic by user99914, dustin, Venus, colormegone, Shaun Jan 20 '15 at 8:22

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  • $\begingroup$ You must divide out some arrangements because of the repeated letters of C and R. $\endgroup$ – turkeyhundt Jan 20 '15 at 6:37
  • $\begingroup$ ya exactly, i m facing the same problem.. $\endgroup$ – Pratyush Jan 20 '15 at 6:38
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An essential point to note is that all orbits under cyclic rearrangement have $8$ distinct elements, since there are some letters that occur only once and can serve as marker. Now one can just count the number of permutations of the letters, and divide by $8$, namely $\frac{8!}{2!2!1!1!1!1!}/8=1260$.

Just to show the contrast with not having a singleton letter as marker, let me also count the cyclically distinct permutations of AAAABBCC. Now there are $\frac{8!}{4!2!2!}=420$ permutations to begin with, for which we need to count the cyclic orbits. Most orbits will have $8$ elements, but some orbits will have only $4$ elements because a cyclic shift of $4$ maps the permutation to itself (even larger symmetry is not possible). The number of permutations with that property is $\frac{4!}{2!1!1!}=12$: they are formed of a permutation of AABC repeated twice. So the solution of this alternative problem is $(420-12)/8+12/4=51+3=54$.

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So, I would say there are $8!$ ways to arrange $8$ letters in a row, so $7!$ ways in a circle. But in this case, divided by $2!$ twice for the repeated $C$ and $R$. So $$\frac{7!}{2!2!}=1260$$

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