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 ________. [2026]

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 ________. [2026]

Smart Achievers · Accepted Answer

(126)

Let B's wins = k

Total games played = 5 + k

Last game is won by A so among the first 4+ k games,

A wins 4 times

B wins k times

$∴$ Number of ways in which A wins = 4 + k

Sum of all possible values of k

$= \sum_{k = 0}^{4} {}^{4 + k}C_{4} + {}^{4}C_{4} + {}^{5}C_{4} + {}^{6}C_{4} + {}^{7}C_{4} + {}^{8}C_{4}$

= 1 + 5 + 15 + 35 + 70 = 126

Solution

Q.
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 ________. [2026]

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Solution

Q. 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 ________. [2026]

Ans. (126) Let B's wins = k Total games played = 5 + k Last game is won by A so among the first 4+ k games, A wins 4 times B wins k times ∴ Number of ways in which A wins = 4 + k Sum of all possible values of k =∑k=04C44+k+C44+C45+C46+C47+C48 = 1 + 5 + 15 + 35 + 70 = 126

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Q.
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 ________. [2026]