In a unique hockey series between India and Pakistan, they decide to play on till a team wins 5 matches. The number of ways in which the series can be won by India, if no match ends in a draw is :
A) 126 (EDIT: earlier it was 125 due to typing error, so in answers question is critisied for that)
B) 252
C)225
D)none
I cannot figure out what to do. Its a question of permutation and combination. The answer is A)126
On
So there can either be 5,6,7,8 or 9 games in the series (if it reaches 9 games, the result has to be 5-4).
5 games: India win every game - 1 combination
6 games: India win all but one - they must have lost 1 of the first 5 games - 5 combinations
7 games: India lose 2 of the first 6 games - ${6}\choose{2}$ combinations - 15 combinations
Continue like this and you should get the correct answer which is 126, not 125.
The series cannot go on for more than $9$ games, since at that point one of the teams must have won $5$ games. It's convenient, mathematically, to imagine that the series continues, even after the winner has been determined, to a full $9$ games, with the winner simply forfeiting the unnecessary games, so that they wind up winning the series 5-4. The number of different ways this can happen is
$${9\choose5}={9!\over5!4!}={9\cdot8\cdot7\cdot6\over4\cdot3\cdot2\cdot1}=126$$