SOLUTION: Imagine a checkers tournament between Person A and Person B (A and B). They play at most 2n matches and whoever first wins n+1 matches wins the tournament (the tournament ends once

Question 440717: Imagine a checkers tournament between Person A and Person B (A and B). They play at most 2n matches and whoever first wins n+1 matches wins the tournament (the tournament ends once one person wins n+1 matches). Two tournaments are different if they have a different sequence of games where A wins and where B wins. Assume that A wins the tournament (so the tournament has n+1 wins for A). How many different tournaments are there where B wins exactly r games (r is less than n)?
Answer by richard1234(7193) (Show Source): You can put this solution on YOUR website!
The tournament can end up with A winning n+1 games, B winning anywhere from 0 to n games. If B wins r games, then we have a sequence of n+1+r games to "arrange" in a line. This can be done using (n+1+r)Cr = (n+1+r)C(n+1) ways. Summing up from r = 0 to r = n:

+ ... + . If you know the hockey stick identity with Pascal's triangle, we can recall that this sum takes that exact same form, and the sum is equal to .
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