How many elements of order 5 are there in $S_7$ This seems relatively simple, since the lowest common multiple of the length of the disjoint cycles must be the order of the permutation. Writing this I got:
$$(5)(1)(1)$$
Is the only cycle in $S_7$ whose lowest common multiple that I could find. The number of ways to write this are $\frac{7!}{5}$, but my book is instead doing this:
$$\frac{7!}{2!\cdot 5}$$
Why is this the right answer for the combinations? And why dividing by $2!$? It seems arbitrary to me, could anyone englighten me on this?
 A: This is really
$$\binom75\frac{5!}{5}$$
First you choose five of the seven numbers to use in the cycle, then you can arrange them in $5!$ ways, but they come in groups of $5$ arrangements that are equivalent by cyclic permutation, hence the division by $5$.
A: There are $\binom{7}{2} = \frac{7!}{5!(7-5)!} = \frac{7!}{5!2!}$ ways to pick $5$ elements ignoring order.  Having chosen the elements in your cycle, there are $5!$ ways to order them, but those ways break into sets of $5$ which are the same except for starting point.  For instance, $(1\,2\,3\,4\,5)$, $(2\,3\,4\,5\,1)$ are listed in distinct orders, but are the same cycle.  So really, there are only $\frac{5!}{5}$ distinct $5$-cycles on a given set of elements.
Therefore, there are $\frac{7!}{5!2!} \cdot \frac{5!}{5} = \frac{7!}{2! \cdot 5}$ such elements of $S_7$.
A: A combinatoric proof for the formula is to use your $(5)(1)(1)$ cycle decomposition.  You can put the elements in order in $7!$ ways. The first five of your order will be the $5-$cycle.  You can rotate the cycle in $5$ ways, which is the division by $5$ you have.  Swapping the order of the last two elements doesn't change the permutation, so that is the division by $2!$, giving the book answer.
