Combinatorics problem. Discrete Mathematics II

[Mei1]

There is a table tennis tournament consisted of 8 participants that is guided by the following rules:

1. Each player plays with every other player for exactly one party
2. If in the i-round there was a party between A and B and a party between C and D, and A and C play In i+1, then in i+1 round B and D have to play too.

In how many different ways can a schedule for all rounds be made, not giving importance to the table on which each player plays?

 

[Greg]

Hello and welcome to MHB, Mei! :D

We ask that our users show their progress (work thus far or thoughts on how to begin) when posting questions. This way our helpers can see where you are stuck or may be going astray and will be able to post the best help possible without potentially making a suggestion which you have already tried, which would waste your time and that of the helper.

Can you post what you have done so far?

A similar post was made http://mathhelpboards.com/advanced-probability-statistics-19/combinations-problem-18376.html.

 

[I like Serena]

There is a table tennis tournament consisted of 8 participants that is guided by the following rules:

1. Each player plays with every other player for exactly one party
2. If in the i-round there was a party between A and B and a party between C and D, and A and C play In i+1, then in i+1 round B and D have to play too.

In how many different ways can a schedule for all rounds be made, not giving importance to the table on which each player plays?

Hi Mei! Welcome to MHB! (Smile)

Let's start with the first round.
We can order the contestants in $8!$ ways.
Since (A,B) is the same game as (B,A), which applies to each pair, we should divide by $2^4$.
Then, since the table it not important, we should also divide by $4!$.
So the number of possible combinations for the first round is:
$$\frac{8!}{2^4 \cdot 4!} = 105$$

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