Help with Combinations Problem: 8 Contestants, 8 Games

[flashback]

Could anyone give me a help with this combinatorics problem:

On a tournament of ping-pong there are 8 contestants and these rules apply:
-Every player plays every other exactly once.
-If in the i-th round there was a match between A and B and a match between C and D, and in the i+1-st round there is a match between A and C, than in the i+1-st game there must be a match between B and D.
How many different ways can a schedule for all games be made, if it does matter which player plays on which table?

 

[Greg]

flashback, you have been advised of http://mathhelpboards.com/rules/ (in particular, rule #11). Do you have any thoughts on how to begin? Have you progressed towards a solution?

A similar post was made http://mathhelpboards.com/discrete-mathematics-set-theory-logic-15/combinatorics-problem-discrete-mathematics-ii-18379.html.

 

[flashback]

As i am new to combinatorics, i do not have an idea how to approach this problem. Because of this i am asking for an advice

 

[flashback]

I am on a clue now, it is i think that there must be no repetition since 2 players can not play two others again because as i understood this is a elimination tournament, but nothing aside from this

 

[I like Serena]

I am on a clue now, it is i think that there must be no repetition since 2 players can not play two others again because as i understood this is a elimination tournament, but nothing aside from this

Hi flashback! (Smile)

I don't think it's an elimination event, since it says that each player plays against each other player exactly once.
Moreover, that means there are 7 rounds of 4 matches each.
For the first round we can order the contestants in $8!$ ways.
I'm assuming that $A$ playing $B$ is the same match as $B$ playing $A$.
That means we should divide by $2^4$ to compensate for the ordering of pairs.
Since the table is apparently important, that means that the number of possibilities for the first round is:
$$\frac{8!}{2^4}= 2520$$

Hmm... what about the second round? (Wondering)

 

[flashback]

Well since it is not an elimination match, then it would be the same as the first one or am i wrong?

 

[flashback]

I am sorry but, i have made a mistake in the question. the table on which a player plays does NOT matter. So i guess we should divide it by 4! since in 8 matches we would have 4! tables which are overcounted right?

 

[I like Serena]

Well since it is not an elimination match, then it would be the same as the first one or am i wrong?

I am sorry but, i have made a mistake in the question. the table on which a player plays does NOT matter. So i guess we should divide it by 4! since in 8 matches we would have 4! tables which are overcounted right?

Ah okay, so this thread is identical to the other one, isn't it?

Anyway, the second round can't be identical to the first one, since the first rule states that everyone plays every other person exactly once.
Suppose the first round is (A,B), (C,D), (E,F), (G,H).
Then the second round cannot contain any of those matches.
So suppose the second round contains (A,C), what does the second rule imply then? (Wondering)

 

[flashback]

well if A and C play in the second round than B and D play in the second round aswell, if i understood ur question correctly. And in addition i think there will be 4 rounds because in that way everyone could play everyone. but i do not know how many combinations go until those rounds. i understood the 1st round part but i can not proceed to the other steps

 

[squidsk]

There are 7 rounds. If you have eight teams for anyone team to play all the others means that each team must play 7 other teams, at one team per round that makes 7 rounds.

 

[I like Serena]

well if A and C play in the second round than B and D play in the second round aswell, if i understood ur question correctly. And in addition i think there will be 4 rounds because in that way everyone could play everyone. but i do not know how many combinations go until those rounds. i understood the 1st round part but i can not proceed to the other steps

Correct!

That leaves (E,F), (G,H) who played in the first round.
Who can they play next? How many possibilities?

 

[flashback]

That is where my problem is. I cannot understand what the next steps are. i am going on thoughts that there are 4!/(2^2*2!) ?

 

[I like Serena]

Correct!

That leaves (E,F), (G,H) who played in the first round.
Who can they play next? How many possibilities?

That is where my problem is. I cannot understand what the next steps are. i am going on thoughts that there are 4!/(2^2*2!) ?

Huh?
Let's take this one step at a time.
E can play G, in which case F will play H.
Or E can play H, in which case F will play G.
Isn't that just 2 possibilities? (Wondering)

 

[flashback]

Well i was in my thoughts on that because everyone plays someone in first step, and in the second step they can play with 2 less combinations more, but i would appreciate if you could explain the full solution ty :)

 

[flashback]

I got somewhat of a solution that in the 2nd round there will be 12 possible solutions, am i somewhat on a good path?