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

Nov 2, 2014

Number of Combinations

Let $C \left(n , r\right)$ 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

$C \left(14 , 2\right) = \frac{14 \cdot 13}{2 \cdot 1} = 91$.

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

$C \left(12 , 4\right) = \frac{12 \cdot 11 \cdot 10 \cdot 9}{4 \cdot 3 \cdot 2 \cdot 1} = 495$.

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

$91 \cdot 495 \cdot 1 = 45045$.

I hope that this was helpful.