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Which of the following numbers can be orders of permutations \alpha of 11 symbols such that it does not fix any symbol?
1. 18,
2. 30,
3. 15,
4. 28

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A permutation $\pi$ can always be written as product of disjoint cycles $$ \pi=c_1\cdot\ldots\cdot c_k $$ and then the order of $\pi$ is the lcm of the lengths of the cycles $c_1,\dots,c_k$.

Thus you are looking for partitions $$ 11=\ell_1+\dots+\ell_k $$ with $\ell_i\geq2$ for all $i$ such that $lcm(\ell_1,...,\ell_k)$ is some specified value.

For instance $11=5+6$ proves that there's a permutation as required of order 30.

Try to work out the other values by yourself!

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yes 18,24,28,30 are the possible orders ,thank you very much. – antara Dec 19 '12 at 14:09
what about $15$??????? – Un Chien Andalou May 31 '13 at 18:03
@TaxiDriver : $11=5+3+3$ and $lcm(5,3,3)=15$. – Andrea Mori Jun 1 '13 at 9:20
thankkuuuuuuuuuuuu – Un Chien Andalou Jun 1 '13 at 9:24

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