SOLUTION: 42.A team plays 15 games a season. How many ways are there to end up with 8 wins and 7 losses for the season?
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Question 1134436: 42.A team plays 15 games a season. How many ways are there to end up with 8 wins and 7 losses for the season?
Answer by mathsolverplus(88) (Show Source): You can put this solution on YOUR website!
This situation assumed that there are only two outcomes and tying doesn't exist.
Let C(n,r) denote the number of combinations of n items chosen r items at a time.
C(n,r) =
Number of ways to win 8 out of 15 games:
C(15,8) =
=
= 6435
Number of ways to lose 7 out of 7 remaining games:
C(7.7) = = 1
Hence, the total number of ways to have 8 wins and 7 losses is:
6435*1 = 6435
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