# How many ways can team $X$ win a hockey series in which teams $X$ and $Y$ play until a team wins $10$ matches?

In a hockey series between $X$ and $Y$, they decide to play until a team wins $10$ matches. What is the number of ways in which team $X$ wins?

My attempt (from the comments):

$9C0 + 10C1 + 11C2......... +19C9$ ...last match must be won by X right?

• Please share what you have tried. This isn't a place to copy & paste your homework without any effort. – Shaun Sep 3 '17 at 12:26
• I tried hard bro....I'm trying it from past 2 days – umang Sep 3 '17 at 12:27
• Yes, but what have you tried? – Shaun Sep 3 '17 at 12:28
• – Shaun Sep 3 '17 at 12:30
• Have another think about the last term & look at this en.wikipedia.org/wiki/Hockey-stick_identity – Donald Splutterwit Sep 3 '17 at 12:45

## 1 Answer

For team $X$ to win the tournament in $k$ matches, it must win $9$ of the first $k - 1$ matches and then win the $k$th match. Notice that $k \leq 19$, for otherwise team $Y$ wins the tournament. Hence, the number of ways $X$ can win is $$\binom{9}{9} + \binom{10}{9} + \binom{11}{9} + \ldots + \binom{18}{9} = \binom{19}{10}$$ by the hockey-stick identity. Notice that the answer is the number of ways team $X$ can win $10$ of the first $19$ matches.