Solution verification: How many elements of order 3 are there in $S_5$

Question

How many elements of order 3 are there in $S_5$.

I found it a several ways to solve this one. I did like this, but do not know is it correct:

I have a possible options (a b c) = ( b c a) = ( c a b) (i), because it is cyclic. So now we have a to chose on 5 ways, b on 4 and c on 3 (ii). Having on mind (i) and (ii) we have :

$$\frac{5*4*3}{3}=20$$. Is this solutin correct?

Answer 1

Yes, it is correct.

I would have done it in another way. First you pick $3$ elements out of $5$; there are $\binom53$ such choices. Once that choice is made, you order them in every possible way; then you get $\binom533!$. Finally, since, as you wrote, $(a\ \ b\ \ c)=(b\ \ c\ \ a)=(c\ \ a\ \ b)$, you have to divide what you got by $3$. So, the final answer is$$\binom53\frac{3!}3=20.$$

Answer 2

The elements of order $3$ are the $3$-cycles.

There are $5\cdot4\cdot3/3=20$. That's $60$ permutations of $3$ out of $5$, divided by $3$ cyclic permutations of each $3$-cycle.