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Professional sports teams commonly end their seasons with a championship between two teams. The series ends when one team has one 4 games and so must last at least 4 games and at most 7. How many different sequences of game winners are there in which the game ends in 5 games? 6 games? 7 games?

My attempt involved noticing that there are two teams, and realizing that if 5 games have been played, one team must have won 4 and the other won 1 game. So I tried seeing how many different ways that one team can win 4 games and the other team can win 1, but I cant seem to get the correct answer.

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  • $\begingroup$ oops, sorry. My attempt involved noticing that there are two teams, and realizing that if 5 games have been played, one team must have won 4 and the other won 1 game. So I tried seeing how many different ways that one team can win 4 games and the other team can win 1, but I cant seem to get the correct answer. $\endgroup$ – 1233211 Dec 29 '15 at 5:11
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    $\begingroup$ Are you forgetting that the eventual winner must win the last game? So if team A wins in five games there are only four sequences (not five) because A can't win the first four. $\endgroup$ – Ross Millikan Dec 29 '15 at 5:24
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There are two ways that the series can end in 4 games: team A wins four straight or team B wins four straight.

To calculate how many ways team A can win 4 games in a series of 5, note first that team A must win the last game, so we must distribute the remaining 3 wins among the first 4 games, which is $\binom{4}{3} = 4$. Multiply this by 2 to gets the total number of possibilities for either team A winning or team B winning the series.

To calculate for a series of 6 games, we proceed similarly: if team A takes the series, it must win the last game, and there are $\binom{5}{3}=10$ ways to distribute the remaining 3 wins among the first 5 games. Then multiply by 2 to the get the possibilities for team A or team B winning the series.

For a series of 7 games, we have $2\binom{6}{3}=40$.

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