How many elements of order $4$ in $S_{7}$? I'm aware that there a questions like this but I would like to check whether my approach is good.
I know the order of disjoint cycles is the product of their lengths. Thus, to get order $4$ with $S_{7}$, there are two options:
Option $(1)$: [cycle of length 4][cycle of length 1][cycle of length 1][cycle of length 1]
Option $(2)$: [cycle of length 2][cycle of length 2][cycle of length 1][cycle of length 1][cycle of length 1]
For option 1 I would say we have $\frac{7 \cdot 6 \cdot 5 \cdot 4}{4}=120$ possible cycles. 
For option 2 I would say we have $\frac{7 \cdot 6}{2}\cdot \frac{5 \cdot 4}{2}=120$ possibilities. 
This gives a total of $240$ possibilities. But I'm in doubt about option 2. Am I counting double? And if so, is dividing by $2$ the solution?
 A: The order of a product of disjoint cycles is the least common multiple of the orders of the cycles.  
Therefore, elements of $S_7$ with order $4$ could have the form 
[cycle of length 4][cycle of length 2][cycle of length 1] or 
[cycle of length 4][cycle of length 1][cycle of length 1][cycle of length 1].
For the former form, there are $7\times 6 \times 5 \times 4$ possibilities for the cycle of length $4$, though divide by $4$ because the cycle could be written starting with any of the $4$ elements chosen, so $7 \times 6 \times 5=210$ possibilities.  Then there are $3 \times 2$ possibilities for the cycle of length $2$, though divide by $2$ because the cycle could be written starting with either of the $2$ elements chosen.  In sum, there are $7\times 6\times 5\times 3=630$ distinct possibilities.
For the latter form, there are as above $7 \times 6 \times 5 =210$ possibilities for the cycle of length $4$, and the cycles of length $1$ are then determined, so $210$ possibilities.
In sum, there are $630+210=840$ elements of order $4$ in $S_7$.
