Number of elements of order $5$ in $S_7$: clarification In finding the number of elements of order $5$ in $S_7$, the symmetric group on $7$ objects, we want to find products of disjoint cycles that partition $\{1,2,3,4,5,6,7\}$ such that the least common multiple of their lengths is $5$.  Since $5$ is prime, this allows only products of the form $(a)(b)(cdefg)$, which is just equal to $(cdefg)$.  Therefore, we want to find all distinct $5$-cycles in $S_7$.  
Choosing our first number we have seven options, followed by six, and so on, and we obtain $(7\cdot 6 \cdot 5 \cdot 4 \cdot 3)=\dfrac{7!}{2!}$.  Then, we note that for any $5$-cycles there are $5$ $5$-cycles equivalent to it (including itself), since $(abcde)$ is equivalent to $(bcdea)$, etc.  Thus, we divide by $5$, yielding $\dfrac{7!}{2!\cdot 5}=504$ elements of order $5$.
I understand this...it makes perfect sense.  What's bothering me is:
Why isn't the number of elements of order five $\binom{7}{5}/5$?  It's probably simple, I'm just not seeing why it doesn't yield what we want.

This is your code:

g:=SymmetricGroup(7);
h:=Filtered(Elements(g),x->Order(x)=5);
Size(h);

The size of $h$ is the same as other theoretical approaches suggested. It is $504$.
