10 children divide into 2 teams each. How many divisions possible? In order to play a game of basketball, $10$ children at a playground divide themselves into 


*

*$2$ teams of $5$ each

*$2$ teams of $x_1$ and $x_2$


How many different divisions are possible? This is not a homework problem. 
In general, I was thinking of something along the lines of 
$$\binom{10}{x_1\ x_2} + \binom{10}{x_2\ x_1}$$
But as this problem is worded, it's not clear to me if this is overcounting because we have $2$ indistinguishable teams.

Textbook says

 A: (1) Here, we have $\displaystyle \binom{9}{4} = \dfrac{9!}{4!5!}$. The point that matters here is that it is only after one child we'll call "A" takes a stance as a self-appointed captain to pick the  rest of his/her team that the groups become distinguishable: A's group, and A's rejects. So Captain "A", has in fact, nine-choose $4$ ways to select the other four children on his/her now-identifiable group: the "A-team". The five unpicked (A-rejected) children become, by default, A-team's competition.
(2) We can use either the binomial coefficient or its equivalent expression as a multinomial coefficient:  $\displaystyle \binom{10}{x_1} = \binom{10}{x_1, x_2} = \dfrac{10!}{x_1!\cdot x_2 !}$, because $x_1 + x_2 = 10$.
A: One answer to the $5$-$5$ problem can be written as $\frac{1}{2}\binom{10}{5}$. This can be done by deliberate double-counting, and dividing by $2$ to compensate.
Another way depends on a fact you did not know: Alicia is one of the $10$ people, and she likes to run things.
There are $\dbinom{9}{4}$ ways for Alicia to choose the people who will be on her team. Thus the number of divisions into two teams of $5$ is $\dbinom{9}{4}$.
It is easy to verify that the two answers are numerically the same.

For "unbalanced" teams of sizes $x_1,x_2$ where $x_1\ne x_2$, the situation is simpler. There are $\dbinom{10}{x_1}$ ways to choose who will be on the team with $x_1$ people, and now we are done.
Only $5$-$5$ requires special consideration.
A: In fact, you need to divide by $2!$ because the teams aren't labelled as $A$ or $B$. Another way to see this. Let's say there is a John in the group. We pick John's teammates, choosing 
$9\choose 4$=126. This also determines the other team.
For $(2)$, with $x_1\neq x_2$,  John will either be a member of a team with $x_1$ members or a team with $x_2$ members. We then pick its team remaining member
$$
{9\choose x_1-1}+{9\choose x_2-1}.
$$
Note that in 
$$
{9\choose x_1-1}+{9\choose x_2-1}={10\choose x_1}.
$$
To see this, note that ${n\choose k}={n\choose n-k}$, so
$$
\begin{align}
{9\choose x_2-1}&={9\choose 9-(x_2-1)}\\
&={9\choose 10-x_2}\\
&={9\choose x_1}
\end{align}
$$
 Then apply Pascal's rule rule
$$
{n\choose k}={n-1\choose k-1}+{n-1\choose k}
$$
to show
$$
\begin{align}
{10\choose x_1}={10-1\choose x_1-1}+{10-1\choose x_1}= {9\choose x_1-1}+{9\choose x_1} = {9\choose x_1-1}+{9\choose x_2-1}
\end{align}
$$
