Excel Problem I just can’t get my head around.

  • Thread starter Thread starter Ant
  • Start date Start date
A

Ant

I am trying to solve a problem that I think Excel should be able to help me
with, if only I knew how!



The problem is this. I am organising an event and expecting around 40-50
people, each person should visit each of the 6 stands with a group of other
guests. After visiting the first stall, each person should then move onto
another stall; however they should not meet anyone at the second stall that
they were with previously at the first. This should then continue until
every one has visited all 6 stalls and not met with anyone they have been
with at a stall before. I expect there to be approx 10 people at any one
stall at a time e.g. 60 people attend / 6 stalls = 10 at a stall.



I first thought this would be a simple problem, start a table in Excel and
find a pattern then copy the pattern down, but the more I try the more
frustrated I get. Any suggestions welcome.
 
Look up combinations and permutations in a college statistics book, or
search for statistics or probability in your available newsgroups.

WDA

end
 
The problem is not solvable.

Imagine person 1 at the first booth during the first visit, in the
company of persons 2-10.

When person 1 moves on to the second visit at the second booth, in order
to have no overlap with persons 2-10, he would have to have 9 companions
from among the people that originally visited booths 3-6 (since those at
booth 2 couldn't stay there).

Getting 9 people from 4 groups guarantees that some of the people will
have been at the same booth with someone else during the first visit.

Another way to look at it is that persons 2-10, in order to not overlap
with person 1, have to be distributed among the other 4 booths (3-6) -
which can't be done with no overlap among themselves.
 
Thanks for that. OK I can see that now, but I still can't get my head
around how to work out the maximum number of combinations before duplication
occurs.
 
As I pointed out, duplication must occur at the second station,
therefore there won't be any combinations before duplication.
 
I don't think this can be done. The ten people who start at stall 1 would
all have to meet the condition that at the next stall they would be be in
the company of anyone who was present at stall 1. If there are only six
stalls, it is obvious that out of the ten people there will be some
duplication at the stall 2. Not all ten can be with nine new people, four
will have to be duplicates.
 
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