Two players A and B play a series of games of | vedclass

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Question

(A) For player $A$ to win the series, they must win the $5$th game. This implies that in the previous $n-1$ games, player $A$ must have won exactly $4

Vedclass · Accepted Answer

$126$

Vedclass · Answer

(A) For player $A$ to win the series, they must win the $5$th game. This implies that in the previous $n-1$ games, player $A$ must have won exactly $4$ games, and player $B$ must have won $n-5$ games.The series can end in $5, 6, 7, 8,$ or $9$ games.If the series ends in $5$ games: $A$ wins $5$ games, $B$ wins $0$. Ways = $\binom{4}{4} = 1$.If the series ends in $6$ games: $A$ wins $4$ of the first $5$ games and the $6$th game. Ways = $\binom{5}{4} = 5$.If the series ends in $7$ games: $A$ wins $4$ of the first $6$ games and the $7$th game. Ways = $\binom{6}{4} = 15$.If the series ends in $8$ games: $A$ wins $4$ of the first $7$ games and the $8$th game. Ways = $\binom{7}{4} = 35$.If the series ends in $9$ games: $A$ wins $4$ of the first $8$ games and the $9$th game. Ways = $\binom{8}{4} = 70$.Total ways = $1 + 5 + 15 + 35 + 70 = 126$.

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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