Number of conjugates of an $m$-cycle $\sigma$ in $S_n$ I am currently reading the following from Dummit and Foote can someone explain to me why the number of conjugates of $\sigma$ is:

$$\frac{n(n\ -\ 1)...(n\ -\ m\ + \ 1)}{m}\ ?$$
Two permutations in $S_{n}$ are conjugate if and only if they have the same cycle type, so really the question is how many $m$-cycles there are in $S_{n}$. Well, choose your $m$ elements from $n$ in $n\choose{m}$ ways. But we can rearrange those $m$ elements in $m!$ ways. Finally, each of our cycles has been represented $m$ times by a cyclic rearrangement, so we divide by $m$. So our final count is $$\frac{n!}{(n-m)!m}$$
as required.
The number of combinations of $n$ distinct things taken $m$ at a time is your starting point, for which the binomial coefficient comes in handy: $n!/m!(n-m)!$ Since each combination will represent more than just one $m$-cycle, for every choice of the 1st spot of an $m$-cycle there will be $m-1$ choices for the 2nd spot, $m-2$ choices for the 3rd spot, etc. You then arrive at $(m-1)! \times n!/m!(n-m)!$