In how many ways can two sports teams consisting of $11$ members each be formed by $23$ players?
MY WORK:
I tried to choose at first $11$ members from $23$ and then again select $11$ from remaining $12$ of them. So I get : $$\binom{23}{11} \times \binom{12}{11}$$ $$=16224936$$
But again, I can do the division of $23$ players into two groups like - $$\binom{23}{22}\times \frac{22!}{2!(11!)^2}$$ $$=8112468$$ Which is half of the previous.... I know that the later one is correct, but where do I go wrong in doing the first one?
If you need to find the number of ordered pairs (2-tuples) of team, then you may use the first method.
If you are asked to find the number of unordered pairs (2-element sets) of teams, then you may use the second method.
Note that the first number equals to the double second one because two teams of a pair can`t be equal. This means that each set $\{\,t_1, t_2\,\}$ corresponds to exactly two 2-tuples $(t_1, t_2)$ and $(t_2, t_1)$.