Hw7 Problem 5

Problem 5

How many elements of order 5 are there in $S_6$?

Solution

Let see this permutation $\begin{pmatrix}1&2&3&4&5&6\\1&3&4&5&6&2\end{pmatrix}=\begin{pmatrix}1\end{pmatrix} \begin{pmatrix}2&3&4&5&6\end{pmatrix}$.
And $|\begin{pmatrix}1&2&3&4&5&6\\1&3&4&5&6&2\end{pmatrix}|=|\begin{pmatrix}2&3&4&5&6\end{pmatrix}|=5$.
So we need the permutation to be like this kind of form.
We make $(1)$ be fixed and change the order of the corresponded number of 2,3,4,5,6.There are 4! possibilities satisfying the condition "order 5"( Any left number could not correspon to itself.2 could correspond to left 4 numbers and 3 could correspond to left 3 numbers and 4 could correspond to left 2 numbers and 5 could correspond to left 1 number and 6 could correspond to left 1 number).
And there are 6 possibilities of the fixed number ——$(1),(2),(3),(4),(5),(6)$.
Then the sum of order 5 of $S_6$ is $6\times 4!=144$.