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

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Solution

$\begin{matrix} L e t C (n, r) d e n o t e t h e n u m b e r o f c o m b i n a t i o n o f n i t e m s c h o s e n r i t e m a t a t i m e . C (n, r) = \frac{n!}{(n - r)! r!} = \frac{n (n - 1) (n - 2) . . . . . . (n - r + 1)}{r!} T h e n u m b e r o f w a y s t o t i e 2 o u t o f 14 g a m e s c a n b e f o u n d b y C (14, 2) = \frac{14.13}{2.1} = 91 T h e n u m b e r o f w a y s t o w a y s t o l o s e 4 o u t o f 12 r e m a i n i n g g a m e s c a n b e f o u n d b y C (12, 4) = \frac{12 \times 11 \times 10 \times 9}{4 \times 3 \times 2 \times 1} = 495 T h e n u m b e r o f w a y s t o w i n 8 o u t o f 8 r e m a i n i n g g a m e s c a n b e f o u n d b y C (8, 8) = \frac{8!}{8!} = 1 H e n c e, t h e t o t a l n u m b e r o f w a y s t o h a v e 8 - 4 - 2 r e c o r d i s 91 \times 495 \times 1 = 45045 \end{matrix}$

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

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