How many ways can 10 people be split into groups of 2 and 3? How many ways can 10 people be split into groups of 2 and 3? the answer says ${5 \choose 2}$... But isn't it the answer if the question were: "How many ways can 5 people be split into groups of 2 and 3?"
 A: 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.
A: 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: $$\binom{10}{2}\binom{8}{2}\binom{6}{2}\binom{4}{2}\frac{1}{5!} = 945$$
Notice we can do this in another way.
$10/2 = 5$ Therefore there are $5$ groups.
We can arrange 10 people in 10! Ways.
If we pick the $n,n+1$ persons of that arrangement Then we picked 5 times groeps of size $2$.
The order of the 5 groups needs to be removed So we correct $10!$ to $\frac{10!}{5!}$. 
The order of the 2 members in every group needs to be removed too , So we correct $\frac{10!}{5!}$ to 
$$ \frac{10!}{5! 2^5} = 945 $$.
This method gives the same result ($945$).
Case 2 can be done in a similar way.
This answer is similar to the accepted one , but I Just wanted to show an alternative way.
Since this alternative way is also an answer and this whole is too long for a comment I made it into an answer.
A: This is equivalent to splitting a group of 10 people into a group of 5, a group of 3, and a group of 2. There are $10!/(5!3!2!)$ ways to do this. Equivalently, you could think of first choosing five people out of the $10$, and then choosing two people out of the $5$. There are $\binom{10}{5}\cdot\binom{5}{2}$ ways to do this. If you compute both methods give you $2520$.
