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When we count the number of elements in a symmetric group, we uses the formula n!. For example, in a symmetric group of 3 elements, there are (123),(132),(1)(23),(2)(13),(3)(12),(1), a total of 6 = 3! of them. Although I can see how this is true in the permutation in combinatorics, I cannot understand why n! gives me the correct number of permutations in a permutation group. As far as I know, (123),(312),(231) are the same in a symmetric group. Thats why I cannot quite understand the formula. My textbook uses the same reasoning as the permutation in combinatorics. Can someone tell me how the formula is derived? Thanks.

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    $\begingroup$ Why do you think the formulas of combinatorics are no longer valid when they are needed in some other field (or in daily life, for that matter)? $\endgroup$ – Christian Blatter Nov 17 '14 at 11:59
  • $\begingroup$ The cycle decomposition isn't very useful for counting permutations. $\endgroup$ – darij grinberg Nov 18 '14 at 6:07
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A permutation $\tau : \{1,\dotsc,n\} \to \{1,\dotsc,n\}$ is specified by

  • any value of $\tau(1)$
  • the value of $\tau(2)$ which lies in $\{1,\dotsc,n\} \setminus \{\tau(1)\}$
  • etc.
  • the value of $\tau(n)$ which lies in $\{1,\dotsc,n\} \setminus \{\tau(1),\dotsc,\tau(n-1)\}$

Thus, there are $n \cdot (n-1) \cdot \dotsc \cdot 1$ possible permutations.

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