$1 \cdot 21$ Team A has $2 m$ people, Team B has $3 m$ people. Now, $14-m$ people are drawn from Team A, and $5 m-11$ people are drawn from Team B to participate in the game. How many people are there in Team A and Team B respectively? How many different ways are there to select the participants for the game?
[Solution] The number of people selected must satisfy the following system of inequalities:
$$
\left\{\begin{array}{l}
0 \leqslant 14-m \leqslant 2 m \\
0 \leqslant 5 m-11 \leqslant 3 m \\
m \in N .
\end{array}\right.
$$

That is,
$$
\begin{array}{l}
\left\{\begin{array}{l}
\frac{14}{3} \leqslant m \leqslant 14 \\
\frac{11}{5} \leqslant m \leqslant \frac{11}{2} \\
m \in N .
\end{array}\right. \\
\frac{14}{3} \leqslant m \leqslant \frac{11}{2}, \quad m \in N .
\end{array}
$$

Thus,
$$
m=5 \text {. }
$$

Therefore, Team A has $2 m=10$ people,
Team B has $3 m=15$ people.

From Team A, $14-m=9$ people are selected, which has $C_{10}^{9}=10$ ways,
From Team B, $5 m-11=14$ people are selected, which has $C_{15}^{14}=15$ ways.
Hence, the total number of ways to select people for the game is
$$
C_{10}^{9} C_{15}^{14}=150
$$

different ways.