Math Is Fun Forum

patbtw

Member

Registered: 2007-03-04

Posts: 1

Puzzle I need help with!!

Hello there! I was wondering if anyone could help me. Im debating with my friend the answer to a puzzle we are trying to work out the answer to! If reagrds soccer macthes. Basically, we are trying to work out how many different accumlative result combinations there are for 8 games of soccer. For example, :  Obviously one combination is that every home team wins. Another combinaiton would be that every away team wins. Another being that all 8 games finish in a tie.

Then for exapmle 7 of the home teams could win but the 8th home team could either lose or tie!! And this could happen with any 8 of the games. I imagine there must be hundreds of combination. Could anyonecome up with an easy way to find the solution to this problem!!!!???

Thanks!!

Toast

Real Member

Registered: 2006-10-08

Posts: 1,321

Re: Puzzle I need help with!!

EDIT: Along with reading my terribly explained post, check this out - lots of cool, easy-to-understand info on this topic: http://www.mathsisfun.com/combinatorics … tions.html

Hmm, this has to do with the multiplication principle...

Lets do this team by team.

For the first teams, there are 3 possible outcomes: The home team wins, ties, or loses.
For the second teams, there are also 3, as well as 3 for the third, fourth etc.

You can try drawing a tree diagram. If you do this for all 8 teams, you should come up with the answer at the end of the tree - this is, however, exhaustive. If you had drawn the tree diagram, however, you may possibly have realised that for one team, there is 3 possibilites, for two teams there are 3^2 = 9, for three there are 3^3 = 27, for four there are 3^4 = 81, for five there are 3^5 = 243...etc
The possibilities are going up by powers of 3.
So at 8 teams, you would have 3×3×3×3×3×3×3×3 = 3^8 = 6561 possibilities.

In general - if event 1 has x outcomes and event 2 has y outcomes and both events are mutually exclusive, then the total possible outcomes is xy.

Let's try another example:
Lauren can spend her night a number of ways. She can see 1 of 5 movies, then go to 1 of 3 restaurants, then watch 1 of 2 movies on TV, then sleep in 1 of 8 positions.

The total possible outcomes from this would be
Event1 × Event2 × Event3 × Event 4
No.PossibleMovies × No.Possible restaurants × No.PossibleTVmovies × No.Possible sleep positions
5 × 3 × 2 × 8 = 240 different possibilities.

Last edited by Toast (2007-03-04 01:32:59)