I need Math help!

I was given this question and can't figure out how to get the answer.

There are 28 Volleyball teams. They pair off and 14 games are being played.
There are only 2 possible outcomes for each game played. Either a home win, or a home loss.

So for the 14 games played, how many different combinations of outcomes can there be?
This has been driving me nuts for 3 days. I tried drawing a circle with 28 points and drawing a single line from each team to every other team but only got 406 combinations. It seems too low.

Please help !!!

Jeff

 

I think this is correct....

There 378 possible pairings in the first game (team 1 can be paired with potentially 27 other teams, team 2 with 26, team 3 with 25, etc. etc.)

In the second game there would only be 364 possible pairings, since 14 pairngs would have been eliminated in game 1.

In the third game only 350 pairings, etc. etc.

That gives you a grand total of 4,018 possible unique pairings for 14 games. Given each pairing can have 2 outcomes, you have the possibility for 8,036 possible unique outcomes in 14 games.

BUT, my logic could certainly be wrong.

 

http://library.thinkquest.org/C006087/english/interactive.shtml

Here is a decent site that will help you.

 

Only 14 teams would have either a home win or a home loss, the other 14 teams would have either an away win or loss!!!!!!!!!!

 

Right, only two possible outcomes for each unique pairing. There would be only 2 possible outcome associated with each of the potential 378 pairings (or 378 possible games) in the first scheduled set of 14 games, making the total number of possible outcomes in game one 756 (either a home win or a home loss for each of the potential 378 games)

At least I think that is right, perhaps I am not understanding the problem ---- I have a headache.

 

There are 2^14 = 16,384 possible combinations of Home Wins and Losses for the 14 games.

Think about it for the case of 3 games.

Let H = Home team wins, L = Home team loses.

So for 3 games, you could have the following combinations:

H,H,H (home team wins each game)
H,H,L
H,L,H
H,L,L
L,H,H
L,H,L
L,L,H
L,L,L (home team loses each game),

or 2*2*2 = 2^3 = 8 combinations for 3 games.


If you know anything about experimental design, think about the combinations being all the design pts of a full, 2 level factorial design with 14 variables (one for each game).


You would have been there awhile drawing each combination by hand.

 

I am not completely sure abut it but I have an idea.

If there was just one game: 2 possible results = 2^1 = 2
Two games, 4 possible results = 2^2 = 4
Three games, 8 possible results = 2 ^3 = 8

.....

.....

.....

Fourteen games, then 2^14 = 16384 posible results

So I think that´s the result, but I would let others confirm it first. I may be perfectly wrong. In fact, most of the times I am wrong.


Regards

 

Oh, Dr. Puffy, I didn´t see your reply until I sent mine.

I guess we have it then right.

 

This calculation does not make allowance for the number of teams and possible pairings.

For example, there are only 3 teams playing 2 games.

possible pairings for game 1 are:
1 vs 2
1 vs 3
2 vs 3

3 possible pairings (and each game has the possibility of a home win or home loss), 2 possible results for each game = 6 possible
Now game 1 just happens to be 1 vs 2

game 2 could only have two possible pairings
1 vs 3
2 vs 3

2 pairings, 2 possible results = 4 possible

So the total number of possible results for 3 teams playing 2 games is 10, not 4

 

I spent some time on this (I made a mistake on my above post, there would be 2 games played in set 1, leaving only one pairing possible in 2 set, making the number of combinations 8, not 10)

I think the correct formula is as follows:

g-1
C = 2(SUM(p-(g*x)x))
x

where
C = combinations
g = number of games
x = 0
p =
t
SUM(t-n)n
n

where
t=number of teams
n = 1

 

If that´s the case, then the number of al possible combinations is:

(14 12) * (2^14 ) = 14!/(14-2)!16,384 = 14 * 13 * 16,384 = 2,981,888


(14 12) is a combinatorial number, of couse.

 

I meant (14 2) of course (combination on 14 elements taken in pairs). The result is correct, anyway. Greetings.

 

<<This calculation does not make allowance for the number of teams and possible pairings.>>

It does not need to. Reread the language of the question:

<<There are 28 Volleyball teams. They pair off and 14 games are being played. There are only 2 possible outcomes for each game played. Either a home win, or a home loss.

So for the 14 games played, how many different combinations of outcomes can there be?>>

28 teams pair off and 14 games are being played. It says ARE being played...it does not ask what combination of pairings could be played. Think about this physically. You have 14 volleyball courts and 2 teams go to each. If team 1 plays 2, it cannot physically play any of the other teams...and neither can team 2.

It reiterates things by saying, "So for the 14 games PLAYED, how many different combinations of outcomes can there be?" This sentence again contrains the problem. And then of course, you
either have a home win or home loss. This leads us back to the 2^14 calculation.

 

Thanks Dr Puffy,

I didn't read the question that way, leave it to a student of the Bible to make things more complicated than they need to be.