How many elements of order 5 are there in $S_7$

Question

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?

Answer 1

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 $5$.

Answer 2

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 $S_7$.

Answer 3

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.