【Example 6】There are 13 teams participating in a sports competition, using a knockout system, where 3 teams can have a bye in the first round. How many essentially different grouping methods are there?
First, determine 3 teams out of 13 to have a bye in the first round, which can be done in $C_{13}^{3}$ ways; then, pair the remaining 10 teams into 5 groups, which can be done in $10!/\left[(2!)^{5} \cdot 5!\right]$ distinct ways. Next, pair these 5 groups with the 3 teams that had a bye (each team as a group) into 4 larger groups, which can be done in $8!/\left[(2!)^{4} \cdot 4!\right]$ distinct ways. Finally, pair these 4 larger groups into 2 zones, which can be done in
$4!/\left[(2!)^{2} \cdot 2!\right]$ distinct ways. Therefore, the total number of distinct ways to group the teams is
$$
\begin{aligned}
C_{13}^{3} & \cdot \frac{10!}{(2!)^{5} \cdot 5!} \cdot \frac{8!}{(2!)^{4} \cdot 4!} \cdot \frac{4!}{(2!)^{2} \cdot 2!} \\
\quad= & C_{13}^{3} \frac{10!\cdot 8!}{2^{12} \cdot 5!}
\end{aligned}
$$

In general, if there are $m\left(2^{n}<m<2^{n+1}\right)$ teams participating in the competition, with $2^{n+1}-m$ teams having a bye in the first round, then by similar analysis, the number of distinct ways to group the teams is
$$
\begin{array}{c}
C_{m}^{2^{n+1}-m} \cdot \frac{\left(2 m-2^{n+1}\right)!}{(2!)^{m-2^{n}} \cdot\left(m-2^{n}\right)!} \cdot \frac{\left(2^{n}\right)!}{2^{2^{n}}} \\
\quad=C_{m}^{2^{2+1}-m} \cdot \frac{\left(2 m-2^{n+1}\right)!}{2^{m} \cdot\left(m-2^{n}\right)!} \cdot\left(2^{n}\right)!
\end{array}
$$