Example 11 (1988 National High School League Question) Team A and Team B each send out 7 players to participate in a Go chess competition in a pre-arranged order. Both sides start with the No. 1 player competing. The loser is eliminated, and the winner then competes with the No. 2 player of the losing side... until all players of one side are eliminated, and the other side wins, forming a competition process. The number of all possible competition processes is
Solution 1 Fill in $2 C_{13}^{7}=3432$. Reason: First consider the match process where Team A wins. At this point, all 7 members of Team B are defeated. Let the $i$-th member of Team B be defeated in the $k_{i}$-th match $(i=1,2, \cdots, 7)$. Since there are at most 13 matches in total, we have $1 \leqslant i_{1}<i_{2}<\cdots<i_{7} \leqslant 13$. Therefore, the match process where Team A wins corresponds one-to-one with the 7-tuple $\left(i_{1}, i_{2}, \cdots, i_{7}\right)$. The 7-tuple $\left(i_{1}, i_{2}, \cdots, i_{7}\right)$ is a selection of 7 numbers from $\{1,2, \cdots, 13\}$, which has $C_{13}^{7}$ ways of selection. Thus, the match process where Team A wins has $C_{13}^{7}$ ways. Similarly, the match process where Team B wins also has $C_{13}^{7}$ ways. Therefore, the total number of different match processes is $2 C_{13}^{7}=2 C_{13}^{6}=3432$.

Solution 2 First consider the match process where Team A wins. Let the $i$-th member of Team A win $x_{i}$ matches $(i=1,2, \cdots, 7)$, then $x_{1}+x_{2}+\cdots+x_{7}=7$. Thus, the match process where Team A wins corresponds one-to-one with the non-negative integer solution set $\left(x_{1}, x_{2}, \cdots, x_{7}\right)$ of the indeterminate equation $x_{1}+x_{2}+\cdots+x_{7}=7$. Since the indeterminate equation $x_{1}+x_{2}+\cdots+x_{7}=7$ has $C_{7+6}^{6}=C_{13}^{6}$ sets of non-negative integer solutions, the match process where Team A wins has $C_{13}^{6}$ ways. The rest is the same as Solution 1.