Hi,
There are 36 people, who will meet on 7 different evenings, and sit at tables for 6. That is 6 tables of 6 people each, every evening for 7 evenings.
Is it possible to arrange the seating so that each person will share a table with 5 new people every evening? So that by the end of the 7th evening, every person has sat with every other person?
Seems to me, it should be possible, because after 7 evenings, each person will have met (7 evenings time 5 new people) = 35 people. But is there some way to use excel to arrange them, or if not possible, then to have excel return an error?
I was thinking set up each evening as a 6 rows by 6 columns square, where each row would be one table. But I'm stumped how to proceed. I'm now able to set up 3 evenings. But those aren't necessarily the correct arrangements that will allow all 7 evenings to work. I think there's only one right answer.
Here's why theoretically seems to me it should work. Each table would have combin(6,2) = 15 unique pairs. There are 36 people so there are combin(36,2) = 630 unique pairs in total. 630 pairs/15 pairs per table = 42 tables. 7 evenings should do it perfectly.
Occurrs to me that this could be a 6/36 lottery question where someone wants to buy 42 tickets which collectively contain every unique pair from 1 to 36.
Using Windows 7, and Excel 2003. Any ideas how to proceed?
...Susan
There are 36 people, who will meet on 7 different evenings, and sit at tables for 6. That is 6 tables of 6 people each, every evening for 7 evenings.
Is it possible to arrange the seating so that each person will share a table with 5 new people every evening? So that by the end of the 7th evening, every person has sat with every other person?
Seems to me, it should be possible, because after 7 evenings, each person will have met (7 evenings time 5 new people) = 35 people. But is there some way to use excel to arrange them, or if not possible, then to have excel return an error?
I was thinking set up each evening as a 6 rows by 6 columns square, where each row would be one table. But I'm stumped how to proceed. I'm now able to set up 3 evenings. But those aren't necessarily the correct arrangements that will allow all 7 evenings to work. I think there's only one right answer.
Here's why theoretically seems to me it should work. Each table would have combin(6,2) = 15 unique pairs. There are 36 people so there are combin(36,2) = 630 unique pairs in total. 630 pairs/15 pairs per table = 42 tables. 7 evenings should do it perfectly.
Occurrs to me that this could be a 6/36 lottery question where someone wants to buy 42 tickets which collectively contain every unique pair from 1 to 36.
Using Windows 7, and Excel 2003. Any ideas how to proceed?
...Susan
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