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?
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 :



$$\frac{1}{3!} \begin{pmatrix}9\\2\end{pmatrix}\begin{pmatrix}7\\2\end{pmatrix}\begin{pmatrix}5\\2\end{pmatrix}2=2\frac{9\times 8\times 7\times 6\times 5 \times 4}{8\times 6}=9\times 7\times 5\times 8$$
Step 1: Choose, out of 9 possible numbers, the first two for the first 2-cycle. (I believe the number of ways you can do this is $\binom{9}{2}$)

Step 2: Choose, out of the remaining 7 possible numbers, the next two for the second 2-cycle. (I believe the number of ways you can do this is $\binom{7}{2}$)

Step 3: Choose, out of the remaining 5 possible numbers, the next two for the third 2-cycle. (I believe the number of ways you can do this is $\binom{5}{2}$)

Step 4: Arrange the last 3 remaining numbers for the 3-cycle. (I believe the number of ways you can do this is $3!$)

[EDIT] Last Step: Take the product of the number of possible combinations. $$\binom{9}{2}\binom{7}{2}\binom{5}{2}(3!)$$

That is the long way to do it. However, if you simplify that, that would just be $$\frac{9!}{(2!)(2!)(2!)}$$.

By the way, $8!$ counts the number of 8-cycles in $S_9$.