Dividing into 2 teams In how many ways can $22$ people be divided into $ 2 $ cricket teams to play each other?
Actual answer : $\large \dfrac{1}{2} \times \dbinom{22}{11}$
My approach : 
Each team consists of $11$ members. Number of ways to select a team of $11$ members = $ \dbinom{22}{11}$
Number of ways by which $22$ people can be divided into $2$ cricket teams = $\dbinom{22}{11} \times 1$ (since the remaining 11 members will automatically fall into the 2nd team).
I appreciate if somebody would be able to explicate the role of $ \large \dfrac{1}{2} $ here.
 A: Another way: It so happens that one of the $22$ people is the Queen, who of course gets to choose the people who will be on her team. This can be done in $\binom{21}{10}$ ways. 
A: This is because when you choose $ \large11 $ people out of $ \large 22 $ people, there is a complementary team formed on the other side, that is, the other $ \large 11 $ people also form a team. So we overcount by a factor of $  \large 2 $, that is, we count every time twice.
For example, let $ \large 1, 2, 3, 4 $ be the group of people to choose from. We can choose $ \large 2 $ players in $ \large \dbinom{4}{2} $ ways but we overcount by a factor of $ \large 2 $. 
Here, if we choose $\large  2 $ teams, we get
(1, 2)
(1, 3)
(1, 4)
(2, 3)
(2, 4)
(3, 4)

The complementary teams in every case are
(1, 2) (3, 4)
(1, 3) (2, 4)
(1, 4) (2, 3)
(2, 3) (1, 4)
(2, 4) (1, 3)
(3, 4) (1, 2)

We notice that we have counted every time twice. For example, $ \large (1, 2) $ and $ \large (3, 4) $ should not be counted separately, as when $ \large (1, 2) $ occurs, we automatically get $ \large (3, 4) $.
We can extrapolate this for the $\large 22 $ players.
