# Finding the number of elements of order 7 in $S_9$

I'm trying to find how many elements in $$S_9$$ have order $$7$$.

Using the fact the order of $$g$$ is the lowest common multiple of the length of its disjoint cycles, the only combination of cycles possible is: one of length $$7$$, and two of length $$1$$.

What I'm having trouble with is finding how many $$g$$ ϵ $$S_9$$ are in this form.

My logic would be you would have $$9$$ options to pick for the "$$1$$ cycle" and then $$8$$ options for the next "$$1$$ cycle" and for the "$$7$$ cycle" there would be $$6!$$ options as each one cannot map to the starting elements.

So the number of elements in $$S_9$$ that have order $$7$$ would be $$9*8*6! =51840$$ yet I'm not sure this is correct because I have seen someone post a general formula for a similar problem (link below) that does not yield the same answer (It actually gives half of my answer, $$25920$$), can someone please point out what I'm missing?

Thanks.

Finding the number of elements of particular order in the symmetric group

The order of the two $$1$$-cycles doesn‘t matter, so you need to divide by $$2!=2$$.