27. Two Wei Qi teams, $A$ and $B$, each comprising of 7 members, take on each other in a competition. The players on each team are fielded in a fixed sequence. The first game is played by the first player of each team. The losing player is eliminated while the winning player stays on to play with the next player of the opposing team. This continues until one team is completely eliminated and the surviving team emerges as the final winner - thus, yielding a possible gaming outcome. Find the total number of possible gaming outcomes.
27. Answer: 3432
We use $a_{1}, a_{2}, \cdots, a_{7}$ and $b_{1}, b_{2}, \cdots, b_{7}$ to denote the players of Team $A$ and Team $B$, respectively. A possible gaming outcome can be represented by a linear sequence of the above 14 terms. For instance, we may have $a_{1} a_{2} b_{1} b_{2} a_{3} b_{3} b_{4} b_{5} a_{4} b_{6} b_{7} a_{5} a_{6} a_{7}$ which indicates player 1 followed by player 2 from Team $A$ were eliminated first, and the third player eliminated was player 1 from team $B$. However Team $A$ emerged the final winner as all seven players of Team $B$ gets eliminated with $a_{5}, a_{6}$ and $a_{7}$ remaining uneliminated. (Note $a_{6}$ and $a_{7}$ never actually played.) Thus, a gaming outcome can be formed by choosing 7 out of 14 possible positions for Team $A$, with the remaining filled by Team $B$ players. Therefore, the total number of gaming outcomes is given by $\binom{14}{7}=\frac{14!}{(71)^{2}}=3432$.