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Mathematics 8 Online
OpenStudy (rsadhvika):

5 teams are participating in a tournment; each team will play every other team; win = 1 point; lose = 0 point

OpenStudy (rsadhvika):

How many total permutations are possible in points table

OpenStudy (rsadhvika):

for example, below is one possible permutation : 4 3 2 1 0 it means : first team : win = 4, lose = 0 second team : win = 3, lose = 1 third team : win = 2, lose = 2 fourth team : win = 1, lose = 3 fifth team : win = 0, lose = 4

OpenStudy (rsadhvika):

help !

OpenStudy (anonymous):

do you have an answer?

OpenStudy (anonymous):

2^5?

OpenStudy (anonymous):

You use combinations for this.

OpenStudy (rsadhvika):

yes

OpenStudy (rsadhvika):

I have been trying from last couple of days

OpenStudy (anonymous):

Hmmm

OpenStudy (rsadhvika):

some combinations/permutations are missing some how how do I know below permutation is not valid ? 4 4 2 0 0 It is not valid, Is there any general method to prune them out ?

OpenStudy (anonymous):

each team plays 4 games...

OpenStudy (rsadhvika):

So more than 1 team cannot win all 4 games

OpenStudy (anonymous):

total wins is 10 and total loses is 10

OpenStudy (rsadhvika):

SUM w = 10 SUM L = 10

OpenStudy (rsadhvika):

each team w <= 4, : <= 4

OpenStudy (rsadhvika):

requires more constraining

OpenStudy (anonymous):

Okay, so here is my plan.... The number of wins for team \(i\) will be \(w_i\).

OpenStudy (rsadhvika):

I am following

OpenStudy (anonymous):

\[ \sum_iw_i = 10 \]

OpenStudy (anonymous):

If we can distribute the wins, then loses will be determined by that.

OpenStudy (rsadhvika):

Yes I got that !

OpenStudy (anonymous):

We can't have 4+4+2+0 or 4+4+0+0 like you said...

OpenStudy (rsadhvika):

however both are different cases 4 + 4 + 0 + 0 is easy to prune out as the sum is not 10

OpenStudy (anonymous):

We could sort the teams by the number of wins, and then find permutations after the fact

OpenStudy (rsadhvika):

4 3 _ _ _

OpenStudy (rsadhvika):

remaining teams must add up to 10-4-3 = 3

OpenStudy (rsadhvika):

will there be some invalid permutations ?

OpenStudy (rsadhvika):

4 3 3 0 0 I think it is invalid

OpenStudy (rsadhvika):

somehow i feel helpless identifying the pattern :/

OpenStudy (anonymous):

I'm not sure if there is a simple solution to this, actually

OpenStudy (anonymous):

Hmmm, ignoring scores for a second, the total number of match ups would be...

OpenStudy (anonymous):

I think 4 3 3 0 0 is invalid.

OpenStudy (anonymous):

The first team never loses. The second team can only lose to the first team. The third team can only lose to the first team. Does the third team win or lose to the second team?

OpenStudy (anonymous):

I tried to create a simulation, but it would take almost a day to run it on 5 teams. http://jsfiddle.net/wio_dude/9Lebe/25/ It doesn't take long with 4 teams though. I wonder if there is any insight to be found from it.

OpenStudy (dan815):

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