Question

Two players A and B play a series of games of badminton. The player who wins 5 games first, wins the series. Assuming that no game ends in a draw, the number of ways in which player A wins the series is _________.

Answer 1

Correct Answer: 126

Step 1: Understanding the Question:
This is a standard problem of finding sequences of wins. For A to win the series, they must win exactly 5 games, and the 5th win must be the last game played.
Step 2: Key Formula or Approach:
If A wins the series in $n$ games, A must have won 4 out of the first $(n-1)$ games and then won the $n$-th game. The number of ways is $\binom{n-1}{4}$. The value of $n$ can range from 5 to 9.
Step 3: Detailed Explanation:
Let $n$ be the total number of games played. For A to win, $5 \le n \le 9$.
- If $n = 5$: A wins all 5. Ways = $\binom{4}{4} = 1$.
- If $n = 6$: A wins 4 of first 5, then wins 6th. Ways = $\binom{5}{4} = 5$.
- If $n = 7$: A wins 4 of first 6, then wins 7th. Ways = $\binom{6}{4} = 15$.
- If $n = 8$: A wins 4 of first 7, then wins 8th. Ways = $\binom{7}{4} = 35$.
- If $n = 9$: A wins 4 of first 8, then wins 9th. Ways = $\binom{8}{4} = 70$.
Total ways = $1 + 5 + 15 + 35 + 70 = 126$.
Step 4: Final Answer:
The total number of ways player A wins the series is 126.