Find the number of elements of order 3 in $S_7$ I understand that there are two cycles of length 3, $(i,j,k)(a,b,c) \in$ $S_7$. 
However, I'm quite stumped in figuring out the logic behind these steps, leading to the answer :
Number of distinct 3 cycles : $(7*6*5)/3$ = 70
Number of distinct elements of the form (i,j,k)(a,b,c) = $(70*((4*3*2)/4))/2$ = 280 distinct elements
I would appreciate any help in terms of understanding the reasoning behind these steps.

There are the 3-cycles:
For each choice of $\{a,b,c\}$ there are two 3-cycles: $(a,b,c)$ and $(a,c,b)$. The total number of them is:
$$
2\times \binom 73 =  \frac{7\times 6\times 5}{3} = 70
$$
Then when there are two 3-cycles: first choose the elements of the first 3-cycle,
then there are 2 3-cycles possible. Then choose the 3 elements of the second 3 cycles, then the 3-cycle. 
Eventually you count every element of order 3 twice (the first and second cycle being both 3-cycle): so divide the final number by 2.
$$
\frac 12\times  \left[2\times\binom 73\right]\times
\left[2\times \binom 43 \right]= 70\times 4 = 280
$$
so there is a total of $350$ elements of order 3.
