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?

Algebra ->  Permutations -> 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?      Log On


   

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) About Me  (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) = n%21%2F%28%28n-r%29%21r%21%29
Number of ways to win 8 out of 15 games:
C(15,8) = 15%21%2F%28%2815-8%29%218%21%29
= %2815%2A14%2A13%2A12%2A11%2A10%2A9%29%2F%287%2A6%2A5%2A4%2A3%2A2%2A1%29
= 6435
Number of ways to lose 7 out of 7 remaining games:
C(7.7) = 7%21%2F7%21 = 1
Hence, the total number of ways to have 8 wins and 7 losses is:
6435*1 = 6435

For more additional instant FREE math help, please check out and follow us on social media!!
Instagram (@mathsolver.plus) : https://www.instagram.com/mathsolver.plus/
Twitter (@MathSolverPlus): https://twitter.com/MathSolverPlus
Snapchat (@mathsolverplus)
Email: mathsolverplus@gmail.com