10. Before the World Cup, the coach of country $F$ plans to evaluate seven players, $A_{1}, A_{2}, \cdots, A_{7}$, by having them play in three training matches (each 90 minutes long). Assume that at any moment during the matches, only one of these players is on the field, and the total playing time (in minutes) of $A_{1}, A_{2}, A_{3}, A_{1}$ can all be divided by 7, while the total playing time (in minutes) of $A_{3}, A_{i i}, A_{7}$ can all be divided by 13. If there are no limits on the number of substitutions per match, how many different scenarios are there for the total playing time of each player?
10. Solution: Let the playing times of $A_{1}, A_{2}, A_{3}, A_{4}$ be $7 k_{1}, 7 k_{2}, 7 k_{3}, 7 k_{4}$ (minutes), with their sum being $x$; the playing times of $A_{5}, A_{6}, A_{7}$ be $13 k_{5}, 13 k_{6}, 13 k_{7}$ (minutes), with their sum being $y$.
Then the problem requires finding the number of positive integer solutions $(k_{1}, k_{2}, \cdots, k_{7})$ that satisfy $7\left(k_{1}+k_{2}+k_{3}+k_{4}\right)+13\left(k_{5}+k_{6}+k_{7}\right)=3 \times 90=270$.
Let $7 x+13 y=270 \Rightarrow x=\frac{270-13 y}{7} \geqslant 4$,
Solving, we get $\left\{\begin{array}{l}x=33 \\ y=3\end{array} ;\left\{\begin{array}{l}x=20 \\ y=10\end{array} ;\left\{\begin{array}{l}x=7 \\ y=17\end{array}\right.\right.\right.$.
Thus, we have the system of equations
$$
\left\{\begin{array}{l}
k_{1}+k_{2}+k_{3}+k_{4}=33 \\
k_{5}+k_{6}+k_{7}=3
\end{array} ;\left\{\begin{array}{l}
k_{1}+k_{2}+k_{3}+k_{4}=20 \\
k_{5}+k_{6}+k_{7}=10
\end{array} ;\left\{\begin{array}{l}
k_{1}+k_{2}+k_{3}+k_{4}=7 \\
k_{5}+k_{6}+k_{7}=17
\end{array}\right.\right.\right.
$$

By the stars and bars method combined with the multiplication principle and the addition principle, we know that
the above system of equations has $C_{32}^{3} C_{2}^{2}+C_{19}^{3} C_{9}^{2}+C_{6}^{3} C_{16}^{2}=42244$ sets of positive integer solutions, and each set of solutions corresponds to a time distribution scenario. Therefore, there are 42244 different scenarios.