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 __________.

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Correct answer: A. A

Player A wins the series if A wins 5 games before B wins 5 games. The series can last for a minimum of 5 games and a maximum of 9 games. If player A wins the series in exactly n games, then A must win the n-th game, and A must win exactly 4 out of the first n-1 games. The number of ways this can happen is given by ^ n-1 C_ 4 . The possible values for n are 5, 6, 7, 8, and 9. Total number of ways for A to win the series is the sum of the number of ways A can win in 5, 6, 7, 8, or 9 games: Total ways = ^ 4 C_ 4 + ^ 5 C_ 4 + ^ 6 C_ 4 + ^ 7 C_ 4 + ^ 8 C_ 4 Using the property ^ n C_ r + ^ n C_ r-1 = ^ n+1 C_ r , we can simplify the sum: ^ 4 C_ 4 = ^ 5 C_ 5 ^ 5 C_ 5 + ^ 5 C_ 4 = ^ 6 C_ 5 ^ 6 C_ 5 + ^ 6 C_ 4 = ^ 7 C_ 5 ^ 7 C_ 5 + ^ 7 C_ 4 = ^ 8 C_ 5 ^ 8 C_ 5 + ^ 8 C_ 4 = ^ 9 C_ 5 Therefore, the total number of ways is: ^ 9 C_ 5 = 9! 5!4! = 9 8 7 6 4 3 2 1 = 126 Answer: 126

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