How to assign 30 persons to rotating 3-member groups

tommyccy

New member
Joined
Jul 31, 2015
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Dear all

I have encountered a real life situation requiring some people good at maths to solve it.
Suppose there are 9 people, identified as 1,2,3,4,5,6,7,8,9
First step : three person as a group
1,2,3
4,5,6
7,8,9
then change partner but if previously as same group cannot be grouped together.
e.g.
1,4,7
2,5,8
3,6,9
then change partner again and if previously as same group cannot be grouped together.

How many groups can be formed and how to arrange these groups.

What will happen if there are 30 persons and 3 persons in a group each time.

The solution will be really useful for me.

Tommy
Thanks
 
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Suppose there are 9 people, identified as 1,2,3,4,5,6,7,8,9
First step : three person as a group
1,2,3
4,5,6
7,8,9
then change partner but if previously as same group cannot be grouped together.
e.g.
1,4,7
2,5,8
3,6,9
then change partner again and if previously as same group cannot be grouped together.
How many groups can be formed and how to arrange these groups.
What will happen if there are 30 persons and 3 persons in a group each time.
It really is impossible to help you because the question is poorly framed.

There are \(\displaystyle \dfrac{9!}{(3!)^3(3!)}\) ways to arrange form three collections of three each.
Note that within each of those "ways" no two subcollections have a common member.

Given thirty people the number of ways to form ten collections of three each is:
\(\displaystyle \dfrac{30!}{(3!)^{10}(10!)}\).

BUT, that may not be at all what you mean???
 
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