In how many ways can the season end with 8 wins, 4 losses, and 2 tie is a college football team plays 14 games?

Question

Ethan Adkins · Accepted Answer

Number of Combinations

Let #C(n,r)# denote the number of combinations of #n# items chosen #r# items at a time.

#C(n,r)={n!}/{(n-r)!r!}={n cdot(n-1)cdot(n-2)cdotcdotscdot(n-r+1)}/{r!}#

The number of ways to tie 2 out of 14 games can be found by

#C(14,2)={14cdot13}/{2cdot1}=91#.

The number of ways to lose 4 out of 12 remaining games can be found by

#C(12,4)={12cdot11cdot10cdot9}/{4cdot3cdot2cdot1}=495#.

The number of ways to win 8 out of 8 remaining games can be found by

#C(8,8)={8!}/{8!}=1#

Hence, the total number of ways to have 8-4-2 record is

#91cdot495cdot1=45045#.

I hope that this was helpful.

Julia Adams · Answer

undefined To calculate the number of ways a college football team can end the season with 8 wins, 4 losses, and 2 ties in 14 games, you can use the binomial coefficient formula. The formula is:

C(n, k) = n! / (k! * (n - k)!)

Where:
- n is the total number of games (14 in this case).
- k is the number of wins (8 in this case).

Using the formula, the calculation would be:

C(14, 8) = 14! / (8! * (14 - 8)!)
         = (14 * 13 * 12 * 11 * 10 * 9) / (8 * 7 * 6 * 5 * 4 * 3 * 2 * 1)
         = 3003

So, there are 3003 ways the season can end with 8 wins, 4 losses, and 2 ties.