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: 126
Solutions · 1
Official solution
For player A to win the series, A must win the 5 game before B does.
So, when the series ends:
A has won exactly 5 games.
B can have won 0,1,2,3, or 4 games.
Now count each case.
Case 1: B wins 0 games
Then the sequence is only
AAAAA
Number of ways =1
Case 2: B wins 1 game
The series has 6 games, and the last game must be won by A.
So in the first 5 games, A must win 4 games and B must win 1 game.
Number of ways
(15)=5
Case 3: B wins 2 games
The series has 7 games, and the last game must be won by A.
So in the first 6 games, A must win 4 games and B must win 2 games.
Number of ways
(26)=15
Case 4: B wins 3 games
The series has 8 games, and the last game must be won by A.
So in the first 7 games, A must win 4 games and B must win 3 games.
Number of ways
(37)=35
Case 5: B wins 4 games
The series has 9 games, and the last game must be won by A.
So in the first 8 games, A must win 4 games and B must win 4 games.
Number of ways
(48)=70
Total number of ways
1+5+15+35+70=126
Hence, the number of ways in which player A wins the series is
126