# Math Help - Pro Football: number of ways to finish with a record of 10-4-2

1. ## Pro Football: number of ways to finish with a record of 10-4-2

As I prepare for finals this coming week, could someone tell me if I have answered the following problem correctly? Please point out any errors and I will attempt to correct the mistakes. Thanks.

The pro football season lasts 16 games. The list WWLTWWWWLWWWLLTW is the record of a team that won its first two games, lost its third, tied its fourth, etc., and finished with a record of 10-4-2, (10wins, 4 losses, 2 ties).

a) How many ways are there for a team to finish 10-4-2?

$\left({16\atop 10,4,2}\right)$ Note: use of multinomial coefficients

b) How many ways in part a) do not have consecutive losses?

First choose the ways to arrange the wins and ties. There are $\left({12\atop 10,2}\right)$ ways to arrange the wins and ties.

Next there are 13 ways to insert a loss into the above sequence. There are now 12 ways to insert the second loss into the sequence with one loss.

There are 11 ways to insert the third loss into the sequence of games. There are 10 ways to insert the fourth loss into the sequence of games. Therefore there are $\left({12\atop 10,2}\right)*13*12*11*10$ ways to play 16 games with 10 wins, 4 losses, and two ties so that there are no consecutive losses.

c) How many ways in part a) have a longest winning streak of six games?

There are 11 ways to place six consecutive wins in the sequence of 16 games. Two of these 11 either start the sequence with a win or end the sequence with a win, so we break into two cases.

First case: the winning streak happens from games 2-15. We have 4 ways to surround this winning streak with a non-win. (2 losses, 2 ties, or a loss before the win streak and a tie after, or a tie before the win streak and a loss after). These translate to the following: losses before and after $9*\left({8\atop 4,2,2}\right)$; ties before and after $9*\left({8\atop 4,4,0}\right)$; a tie before and loss after $9*\left({8\atop 4,3,1}\right)$; and a loss before and tie after $9*\left({8\atop 4,3,1}\right)$. If the season starts with the win streak or ends with the win streak and has a loss on one end: $2*\left({9\atop 4,3,2}\right)$ or has a tie on one end: $2*\left({9\atop 4,4,1}\right)$. There are therefore $9*\left({8\atop 4,2,2}\right)+9*\left({8\atop 4,4,0}\right)+9*\left({8\atop 4,3,1}\right)+9*\left({8\atop 4,3,1}\right)+2*\left({9\atop 4,3,2}\right)+2*\left({9\atop 4,4,1}\right)$ ways to have a longest win streak of six games.

2. Let's do this one at a time.
Part a) is correct.

But in part b) you are treating losses as if they are different 'letters'.
$\displaystyle\binom{12}{10,2}\binom{13}{4}$.

Why? Where does the 13 come from?

3. Originally Posted by Plato
Let's do this one at a time.
Part a) is correct.

But in part b) you are treating losses as if they are different 'letters'.
$\displaystyle\binom{12}{10,2}\binom{13}{4}$.

Why? Where does the 13 come from?
My thinking was to first find the number of ways to arrange the wins and ties. For each of these arrangements, we can insert a loss as the first game played, or the second game played, etc. all the way to the 13th game played. So that is where the 13 comes from. Now we have $\displaystyle\binom{12}{10,2}*13$ arrangements including the first loss.

We want to include the second loss in any place except next to the first loss. There should be 12 ways to include the second loss each of which will give us another arrangement. There should be 11 ways to include the third loss that would not be next to one of the first two losses in the sequence and 10 ways to include the fourth loss. I first drew an array of 12 slots. To include the first loss, it could go before the first or after the last or between any of the first twelve games. Is this reasoning incorrect? I have made many incorrect conclusions this past month or so, but have learned much from each one.

4. Hello, oldguynewstudent!

The pro football season lasts 16 games.
The list WWLTWWWWLWWWLLTW is the record of a team
that won its first two games, lost its third, tied its fourth, etc.
and finished with a record of 10-4-2, (10 wins, 4 losses, 2 ties).

a) How many ways are there for a team to finish 10-4-2?

$\displaystyle{{16\choose10,4,2}}$ . Correct!

b) How many ways in part a) do not have consecutive losses?

First choose the ways to arrange the 10 wins and 2 ties.
There are $\displaystyle{{12\choose 10,2}}$ ways to arrange the wins and ties. . Right!

I agree with Plato . . .

We have (for example):

. . $\_\;W\;\_\;W\;\_\;W\;\_\;W\;\_\:W \;\_\;W\;\_\;W\;\_\;W\;\_\;W\;\_\;W\;\_\;T\;\_\;T\ ;\_$

There are 13 spaces in which place the 4 losses:

. . There are: . $\displaystyle{{13\choose4}}$ choices.

Think about it . . . We can't number the losses.

The first loss $(L_1)$ does not have 13 choices.

Can $L_1$ be in the twelfth space?
. . No! . . . Then it wouldn't be $L_1$ (the first loss), would it?

5. I see this now. I didn't understand Plato. I thought he just wanted more of an explanation on how I got my numbers.

It would be distributing 4 identical losses to 13 distinct slots!

Thanks