two players A and B play a series of games of chess. The winning players in any games gets 1 points while the losing plays get 0 points. The player who achievers 4 point first , wins the series. If no game ends in draw, find the Number of ways in which the series is can be won by A

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Solution

A wins the series the following ways. In all these cases the last match has been win by $A$
$(a)$ A wins all the games in a row. The number of ways. $= 1$
$(b)$ A wins the last game on the remaining $4$ games played. A wins three games while $B$ wins one.
The no. of ways $. \frac{4!}{3!}$
$. \frac{4 \times 3!}{3!} .4$
$(c)$ A wins the last game on the remaining $5$ games played. A wins three while $B$ wins two.
The number of ways $. \frac{5!}{3! . 2!} = 10$ .
$(d)$ A wins the last game on the remaining $6$ games played. A wins three while $B$ wins three.
The number of ways $. \frac{5!}{3! . 3!} = 20$
$∴$ The total no. of ways in which the series can be won by $A = 1 + 4 + 10 + 20 = 35$

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