We can have the groups assigned with the following number of people:
$$(1)\,\{2,2,2,2,2\}\\(2)\,\{2,2,3,3\}$$
Case 1: Choose groups of $2$ from $10, 8, 6, 4$ in succession until we reach our final pair. Our $5$ sets of groups are the same irrespective of the order in which they are chosen, so the number of ways here would be: $$\binom{10}{2}\binom{8}{2}\binom{6}{2}\binom{4}{2}\frac{1}{5!}$$
Case 2:
- Choose $3$ people from $10$.
- Choose $3$ people from the remaining $7$.
- Choose $2$ people from the remaining $4$, leaving the remainder to form the final pair.
In summary, we have $2$ different types of groups, each containing $2$ groups. Eliminating repeats by accounting for the order in which such groups can be chosen, the number of ways here is: $$\binom{10}{3}\binom{7}{3}\binom{4}{2}\left(\frac{4!}{2!\,2!}\right)^{-1}$$
Add the cases together and you have your result.
If the groups are labeled, your answer is correct. Otherwise, notice that choosing five people to be in the first group automatically determines who is in the second group. When you count combinations of five of the ten people, you count each group twice, once when you select the group and once when you select its complement. Therefore, unless the groups are labeled, the number of ways to split ten people into two groups of five is $$\frac{1}{2}\binom{10}{5}$$
Edit: @GlenO proposed the following alternative argument in the comments: Suppose Alicia is one of the ten people. The number of ways two groups of five people can be formed is the number of ways of selecting four of the remaining nine people to be in Alicia's group or, equivalently, the number of ways of selecting five of the remaining nine people to not be in Alicia's group. Therefore, the number of ways to split two people into two groups of five is
$$\binom{9}{4} = \binom{9}{5}$$
which is equivalent to the answer I obtained above.
Best Answer
Looks good to me. As long as you have the Bell numbers available, and aren't expected to derive the 8th Bell number yourself.