Each of two teams has seven players numbered 1 to 7. In the first game, the two players numbered 1 play each other. The loser of each game is eliminated and replaced by the next player of the same team, until all players from one team have been eliminated. Let $N$ be the number of possible sequences of games. Find the remainder when $N$ is divided by 1000.
I listed out possible outcomes when there was 1 player, 2 player, … and the number was 2^something but I got 2097152 and I think I'm doing something wrong.
Best Answer
This is the same as the number of combinations leading to victory in a best-of-$13$ series that ends when one team has clinched the series. Let's assume team $A$ is the winner (and we must remember to multiply our final result by $2$ to account for the possibility that $B$ is the winner).
$A$ can win exactly $6$ of the first $k+6$ games, and then Game $k+7$ (ending the series in exactly $k+7$ games), in $\binom {k+6}{6}=\binom{k+6}{k}$ ways, so we want
$$2\sum_{k=0}^6 \binom {k+6}{k}.$$
This is $2 \cdot (1+7+28+84+210+462+924)= 3432$.
There's another way to get this answer. Each combination of $14$ (not $13$) games resulting in exactly $7$ wins for each team yields a unique combination fitting the conditions, so there are $\binom {14}{7}$ possible combinations. To see that this is a bijection, given such a combination, cut it off when the first team reaches $7$ wins and you get a permissible series. Conversely, given a permissible series, allow the losing team to win all of the remaining (unplayed because superfluous) games, and you end up with a $7$-$7$ tie.