Number of conjugates of $(12)(34)(56)(789) \in S_{10}$
This is how I calculated it and got $840$ as a result:
$${10\choose 2}\cdot{\frac{2!}{2\cdot3!}}\cdot {8\choose 3}\cdot\frac{3!}{3}$$
What I am doing: First I choose two numbers from $10$ (because we have a $2$-cycle present). I multiply by $\frac{2!}{2}$ to account for repeated ones. Further I divide this by $3!,$ since there are three $2$-cycles present. Then I choose three numbers from remaining $8$ and then multiply by $\frac{3!}{3}$ to account for repeated ones. Since there is only a single $3$-cycle present, I don't have to do anything further.
EDIT:-
I think I should be doing this instead :
$$\left\{\left\{{10\choose 2}\cdot\frac{2!}{2}\cdot{8\choose 2}\cdot \frac{2!}{2}\cdot{6\choose2}\cdot\frac{2!}{2}\right\}\frac{1}{3!}\right\}\cdot{4\choose 3}\cdot\frac{3!}{3}=25200$$
What you put in your edit is correct.
You chose two from ten, then accounted for repeats; chose two from the remaining eight, then, again, accounted for repeats; and then chose two from the remaining six, then, once more, accounted for repeats; but these choices could come in any order as they are disjoint: hence the division by $3!$; then from the remaining four elements of the set $S_{10}$ acts on, you choose three elements for the three cycle and then accounted correctly for the repeats up to cyclic permutation.
This all holds because conjugation preserves cycle types.