How many permutations in $S_9$ have one cycle of length $4$, one of length $3$, and one of length $2$?
My attempt so far is: ${9 \choose 4}*{5 \choose 3}*{2 \choose 2}$, which is the way to choose elements in each cycle but then how do I find the numbers ways that the elements can give distinct cycles? Because ${9 \choose 4}*{5 \choose 3}*{2 \choose 2}=1260$ and think the answer should be a lot higher!
If anyone can help me understand what I need to multiply and why!
Thank you!
For the cycle of length $4$, you have $9$ choices for the first element, $8$ for the second, $7$ for the third, and $6$ for the fourth. But you have to divide by $4$ because for example $(1234)=(2341)$.
For the cycle of length $3$ you then have $5$ choices for the first element, $4$ for the second, and $3$ for the third; divide by $3$ because there are three ways to represent the same cycle.
Once you have chosen the cycles of length $4$ and $3$, the cycle of length $2$ is uniquely determined. (There are two possibilities for the first element, but the cycle is the same no matter which of the remaining two elements is written first.)