How many possible ways of organizing the first round ($5$ matches) are there? 
$10$ players play table tennis. How many possible ways of organizing the first round ($5$ matches) are there?

My attempt:
we can choose 2 players form 10 in the first match and 2 players from 8 (expect the first 2 players), and 2 players from 6 (expect the first 4 players)etc...
So the answer should be $\binom{10}{2}\times \binom{8}{2}\times \binom{6}{2}\times \binom{4}{2}\times \binom{2}{2}=113400 $
But the right answer is: $9\times7\times5\times3\times1=945$ which I don't understand
and I don't understand where is the mistake in my solution
I need an explanation to my mistake and the right solution please
 A: Double Factorial
Number the players from $1$ to $2n$. Player $1$ can play $2n-1$ people. For each of those choices, the next lowest unpaired player can play $2n-3$ people. For each of the choices so far, the next lowest unpaired player can play $2n-5$ people. Etc. Thus, there are
$$
\begin{align}
(2n-1)(2n-3)(2n-5)\cdots1
&=(2n-1)!!\\
&=\frac{(2n)!}{2^nn!}
\end{align}
$$
ways to arrange the first round. See Double Factorial.

Overcounting Using Binomial Coefficients
If we count the pairs that can be formed from $10$ people, $\binom{10}{2}$ and then the number of pairs that can be formed from $8$ people, $\binom{8}{2}$, etc.,  we have counted the arrangement
$$
\{(1,2),(3,4),(5,6),(7,8),(9,10)\}
$$
once for each permutation of the pairs. That is $5!$ times. Note that
$$
\begin{align}
\binom{10}{2}\binom{8}{2}\binom{6}{2}\binom{4}{2}\binom{2}{2}
&=\color{#C00000}{\frac{10}2}\cdot9\cdot\color{#C00000}{\frac{8}2}\cdot7\cdot\color{#C00000}{\frac{6}2}\cdot5\cdot\color{#C00000}{\frac{4}2}\cdot3\cdot\color{#C00000}{\frac{2}2}\cdot1\\
&=\color{#C00000}{5}\cdot\color{#C00000}{4}\cdot\color{#C00000}{3}\cdot\color{#C00000}{2}\cdot\color{#C00000}{1}\cdot9\cdot7\cdot5\cdot3\cdot1\\
&=\color{#C00000}{5!}\cdot9!!
\end{align}
$$
That is, $5!$ times the correct number.
