TITLE: How many ways can team $X$ win a hockey series in which teams $X$ and $Y$ play until a team wins $10$ matches?
QUESTION [5 upvotes]: 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?

REPLY [3 votes]: 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.