Teams A and B are playing a series of games. The first team to win 4 games wins the championship. Given that A won the first and second games, how many different sequences of wins and losses are possible until the championship is decided? Assume there are no ties.

1. Problem Description

Teams A and B are playing a series of games. The first team to win 4 games wins the championship. Given that A won the first and second games, how many different sequences of wins and losses are possible until the championship is decided? Assume there are no ties.

2. Solution Steps

Let A represent a win for team A, and B represent a win for team B. Since A has already won the first two games, we need to consider the different ways A can win 2 more games or B can win 4 games.

Since A has already won 2 games, A needs 2 more wins to win the championship, and B needs 4 wins to win the championship.

We can represent the possible sequences of games as follows:

If A wins in the next two games: AA. Then the sequence is AA. Number of ways:

1. Total number of games:

4. If A wins with 3 games: AAB or ABA or BAA.

If A wins with 4 games: AABB, ABAB, ABBA, BAAB, BABA, BBAA

If A wins with 5 games: AABBB, ABABB, ABBAB, ABBBA, BAABB, BABAB, BABBA, BBAAB, BBABA, BBBAA.

Another way to consider this is to look at how many games it takes for A or B to win.

The fewest number of games is 4 (A wins the next two).

The maximum number of games is 7 (B wins 3 games before A wins his 4th).

Let's consider how A can win. A needs to win 2 more games.

If A wins in 2 more games, then A wins the 3rd and 4th games. The sequence of wins is AA.

If A wins in 3 more games, B must win 1 game. A wins the last game. B can win at the 3rd, 4th or 5th game. The sequence is BAA or ABA or AAB. Number of games:

5. If A wins in 4 more games, B must win 2 games. A wins the last game. The two Bs can be arranged in $\binom{3}{2} = \frac{3!}{2!1!} = 3$ ways.

For example, BBAA, BABA, ABBA

However, the number of wins A needs to win is always

4. And the number of wins B needs to win is

4.

If A wins in 4 games, A has 2 wins already, so we need to add two A's and B can win a maximum of 3 games.

Let $n$ be the total number of games played.

$4 \le n \le 7$.

Number of games required to win is

4. If total games played is 4, then A wins games 3,

4. There is 1 sequence, AA

If total games played is 5, B wins 1 game. The sequence ends in A, and one B is in the 3 games before. ${3 \choose 1} = 3$.

If total games played is 6, B wins 2 games. The sequence ends in A, and two Bs are in the 4 games before. ${4 \choose 2} = 6$.

If total games played is 7, B wins 3 games. The sequence ends in A, and three Bs are in the 5 games before. ${5 \choose 3} = 10$.

$1 + 3 + 6 + 10 = 20$

3. Final Answer

20