Question

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

Answer 1

This type of problem is based on the concept of combination. First, we have to consider the total wins in the total number of matches. Then use the combination formula ${}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$. Here, n=14 and r=8. Similarly, we have to find the combination of 4 losses and 2 ties using the formula ${}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$ where n=14, r=4 and n=14, r=2 respectively. Then we have to find the number of possible ways where the season ends with 8 wins, 4 losses and 2 ties. Thus we have to multiply all the combinations and obtain the final answer.

Complete step by step answer:
According to the question, we are asked to find the total number of ways the season end with 8 wins, 4 losses and 2 ties in 14 games.
We have been given that there are 14 games.
We first have to find the total number of ways of 2 ties in 14 games.
We have to use the formula of combinations to find the value, that is,
${}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$.
Here, n is the total number of games and r is the total number of ties.
Therefore, n=14 and r=2.
${}^{14}{{C}_{2}}=\dfrac{\left( 14 \right)!}{2!\left( 14-2 \right)!}$
On further simplification, we get,
${}^{14}{{C}_{2}}=\dfrac{\left( 14 \right)!}{2!12!}$ ----(1)
We know that $n!=n.\left( n-1 \right).\left( n-2 \right)......3.2.1$
Using this factorial formula in expression (1), we get,
${}^{14}{{C}_{2}}=\dfrac{14.13.12.11.10.9.8.7.6.5.4.3.2.1}{\left( 2.1 \right)\left( 12.11.10.9.8.7.6.5.4.3.2.1 \right)}$
Cancelling out the common terms, we get,
${}^{14}{{C}_{2}}=\dfrac{14.13}{2.1}$
On further simplifications, we get,
${}^{14}{{C}_{2}}=\dfrac{14.13}{2}$
$\Rightarrow {}^{14}{{C}_{2}}=7.13$
$\therefore {}^{14}{{C}_{2}}=91$
Therefore, the number of ways of 2 ties in 14 games is 91.

Now let us consider 4 losses.
Since two games are tied, the total number of games for 4 losses is 14-2=12.
To find the number of ways for 4 losses, we have to use the formula of combinations to find the value, that is,
${}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$.
Here, n is the total number of games and r is the total number of losses.
Therefore, n=12 and r=4.
${}^{12}{{C}_{4}}=\dfrac{\left( 12 \right)!}{4!\left( 12-4 \right)!}$
On further simplification, we get,
${}^{12}{{C}_{4}}=\dfrac{\left( 12 \right)!}{4!8!}$ ----(1)
We know that $n!=n.\left( n-1 \right).\left( n-2 \right)......3.2.1$
Using this factorial formula in expression (1), we get,
${}^{12}{{C}_{4}}=\dfrac{12.11.10.9.8.7.6.5.4.3.2.1}{\left( 4.3.2.1 \right)\left( 8.7.6.5.4.3.2.1 \right)}$
Cancelling out the common terms, we get,
${}^{12}{{C}_{4}}=\dfrac{12.11.10.9}{4.3.2.1}$
On further simplifications, we get,
${}^{12}{{C}_{4}}=\dfrac{11.10.9}{2}$
$\Rightarrow {}^{12}{{C}_{4}}=11.5.9$
$\therefore {}^{12}{{C}_{4}}=495$
Therefore, the number of 4 losses in 12 games is 495.

Now let us consider 8 wins.
Since two games are tied and 4 losses, the total number of games for 8 wins is 14-2-4=8.
To find the number of ways for 4 wins, we have to use the formula of combinations to find the value, that is,
${}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$.
Here, n is the total number of games and r is the total number of wins.
Therefore, n=8 and r=8.
${}^{8}{{C}_{8}}=\dfrac{8!}{8!\left( 8-8 \right)!}$
On further simplification, we get,
${}^{8}{{C}_{8}}=\dfrac{8!}{0!8!}$ ----(1)
We know that $n!=n.\left( n-1 \right).\left( n-2 \right)......3.2.1$
Cancelling out the common terms, we get,
${}^{8}{{C}_{8}}=\dfrac{1}{0!}$
We know that $0!=1$, we get,
${}^{8}{{C}_{8}}=\dfrac{1}{1}$
$\therefore {}^{8}{{C}_{8}}=1$
Therefore, the number of ways of 8 wins in 8 games is 1.
Now, we have to find the total number of ways the season ends with 8 wins, 4 losses and 2 ties in 14 games.
That is, we have to multiply all the combinations obtained for 8 wins, 4 losses and 2 ties.
Therefore,
Total number of ways = ${}^{14}{{C}_{2}}\times {}^{12}{{C}_{4}}\times {}^{8}{{C}_{8}}$.
Substituting all the obtained values, we get,
Total number of ways = $91\times 495\times 1=45045$.

Hence, the total number of ways the season ends with 8 wins, 4 losses and 2 ties in 14 games is 45045.

Note: Whenever you get this type of problems, we should use a combination formula. We should not get confused with the permutation formula. We should avoid calculation mistakes based on sign conventions. We can use the below properties of combination for easy calculations:
${}^{n}{{C}_{r}}={}^{n}{{C}_{n-r}},{}^{n}{{C}_{0}}=1,{}^{n}{{C}_{n}}=1$ and ${}^{n}{{C}_{1}}=n$.