<s>[INST] Finding the number of permutations in$S_9$ of the form $(a_1a_2)(a_3a_4)(a_5a_6)(a_7a_8a_9)$ How many permutations, ρ, are there in $S_9$(the group of permutations of order 9!) whose decomposition into disjoint
cycles consists of three 2-cycles (transpositions) and one 3-cycle? In other words, how
many permutations are there in $S_9$ whose decomposition into disjoint cycles is of the form
$(a_1a_2)(a_3a_4)(a_5a_6)(a_7a_8a_9)$?
would I be able to assume that,  $a_1 = 1$, so $a_2$ would have 8 possibile values, $a_3$ would have 7 and so on, for a total of $8!$ such permutations?
 [/INST] *

*You need to choose the support of the first transposition : $\begin{pmatrix}9\\2\end{pmatrix}$ choices.

*You need to choose the support of the second transposition in the remaining set : $\begin{pmatrix}7\\2\end{pmatrix}$ choices.

*You need to choose the support of the third transposition in the remaining set : $\begin{pmatrix}5\\2\end{pmatrix}$ choices.


*You need to divide this number by the number of ways to order the three transpositions (since they all give the same permutation) e.g. : $(1,2)(3,4)(5,6)=(3,4)(1,2)(5,6)$. Hence you need to divide by $3!$.

*Choose the support of the $3$-cycle in the remaining set $\begin{pmatrix}3\\3\end{pmatrix}=1$ and on a support of size $n$ you know that you have exactly $(n-1)!$ different $n$-cycles, here this gives $(3-1)!=2$ choices. 

*Gather everything :
$$\frac{1}{3!} \begin{pmatrix}9\\2\end{pmatrix}\begin{pmatrix}7\\2\end{pmatrix}\begin{pmatrix}5\\2\end{pmatrix}2=2\frac{9\times 8\times 7\times 6\times 5 \times 4}{8\times 6}=9\times 7\times 5\times 8$$
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