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Two teams of 7 players each participate in a tournament. First player of a team plays with the first player of the other team and the loser is eliminated. The winner then player with the next player of the other team. One team wins when the other team is eliminated. A team wins when all the players of the other team are eliminated. In how many ways can the tournament happen?

The winning team cannot lose more than 6 matches. Considering it loses 6 matches of 13, then considering it loses 5 of 12.....until 0 of 7. Is this correct?(multiplied by 2)

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  • $\begingroup$ All the players must be eliminated $\endgroup$
    – user585246
    Aug 18 '18 at 14:43
  • $\begingroup$ Presumably all the players of one team need to be eliminated. Yes to your addition. Now if there are $13$ matches how many ways are there to have them split $7$ wins and $6$ losses for one team? Note that you need to not stop before $13$ for this piece of the count. What does that require? $\endgroup$ Aug 18 '18 at 15:04
  • $\begingroup$ 13 c 7 or 13 c 6 $\endgroup$
    – user585246
    Aug 18 '18 at 15:06
  • $\begingroup$ No, that misses the point of not finishing early. One of your $13 c 7$ is seven wins for the first team to start the tournament. At that point we are done. There is a cute trick that saves you from considering shorter matches and comes out where you are thinking. $\endgroup$ Aug 18 '18 at 15:08
  • $\begingroup$ I'm sorry, I don't understand. $\endgroup$
    – user585246
    Aug 18 '18 at 15:14
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You can just think of it as a match between team A and B with each game counting one point and the winner being the first to seven. If you want the number of ways team A can win in nine games, it must win the last one for the match to end then and you can pick any six of the previous eight for them to win. That says there are ${8 \choose 6}=28$ ways for A to win in nine games. For all the possible lengths we would have $${6 \choose 6}+{7 \choose 6}+{8 \choose 6}+{9 \choose 6}+{10 \choose 6}+{11 \choose 6}+{12 \choose 6}$$ ways for A to win and the same number of ways for B to win.

Rather than evaluate the whole sum, we can just imagine that any match shorter than $13$ games is extended to $13$ games by having the side that lost the match win the remaining games. This makes a bijection between the winning matches and the number of ways to choose the seven games the winning side must win, so there are ${13 \choose 7}=1716$ ways for A to win and the same number for B to win, for a total of $3432$.

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