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

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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) About Me  (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:

S+=+sum%28%28n%2B1%2Br%29Cr%2C+r=0%2C+n%29+=+%28n%2B1%29C0+%2B+%28n%2B2%29C1 + ... + %282n%2B1%29Cn. 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 %282n%2B2%29Cn+=+%282n%2B2%29%21%2F%28%28n%2B2%29%21n%21%29.