Request for help - math problem using binary code

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The discussion centers on a combinatorial problem involving a coffee club of 8 members who meet thrice weekly. The challenge is to determine if it is possible for each member to sit inside and outside at least once, while also sitting at the same table with every other member. The consensus among participants is that achieving this is impossible due to the asymmetry in seating arrangements and the constraints of the pigeonhole principle. The use of binary code was suggested as a potential method for solving the problem, although participants expressed uncertainty about its application.

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Hello

I have math question that I am struggling with that apparently can be solved using code (or reasoning - I tried that and mine didn't work. GO figure :confused: ). Can anyone help me out?

Here is the question:

A coffee club with 8 membres meet every MOnday Wednesday and Friday. At eahc meeting, 4 people sit inside at a table and 4 people sit outside. IS it possible in the course of a week, for each member to be inside at least once, outside at least once, and be at the same table at least once with wach of the other 7 members?

Thanks for the help
 
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Nice question :) But ithink it is impossible because:
Let's name this eight people from 1 to 8

IN OUT
First day: 1234 5678 - A Standard Beginning(Changing places won't make a difference as numbers can be changed too...)


Second D: Assume that we need to make only 2 people(In my sol. they are 1 and 5) meet each people and go both inside and outside, we would just replace them to meet most people and this would be 5234(I)-1678(O)

Third D.: Now we must make 1 and 5 meet each other and let's put them like this
1537 - 2468(There are two contradictions 1st: Its asymmetric, it can't be a solution, there is no sense in putting 1 and 5 to inside not outside(symmetry in solutions for problems like this is VERY Important) and 2nd: 2-7, 4-7, 3-6, 3-8 haven't met)


Another solution trial:(two-to-two meeting system-doesnt works again)
(By this System, we make two people groups like 12, 34... and have only 4 elements)
1st Day:1234(I)-5678(O)
2nd Day:1256(I)-3478(O) (Now we have just 12-78 and 34-56 left)
3rd Day: 3456(I)-1278(O) (All people met; however 7 and 8 has never gone inside)
(The weird thing here is there is no symmetry(Why 7 and 8) and this makes me think I am doing a mistake somewhere...)
(Any explanations to this would be great-just mail me if you have any comments)
 
I tried exactly what you tried, and to no avail. (switching two people at a time and moving one person at a time) I feel like it may be impossible, but by the way my prof was talking in class, it sounded like there was a solution! Hmmm...
He did mention using binary code and pigeon hole principle but I am at a loss as to how to use those (both pretty knew concepts for me, so don't fully understand how to apply to problem!)
Thank you for you help!
 

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