Combinatorics - group rotation Anyone able to provide me with a solution to this problem?
I came across this website whilst struggling with the following problem.  Whilst I have found and read bits that have helped, I still can't solve this particular situation.  I'm not a mathematician in any sense of the word so really would appreciate some help.
I have 5 groups of people.  4 of the groups have 4 people and the other one has 2 people - so 18 people in total.  I have to come up with a system to allow each of the people in each group to meet each other.  The meetings need to be in groups of 3-4.  Some quick and basic maths suggested to me to have 6 meetings of 3 people for 6 weeks.  I just can't come up with the right algorithm to get this organised.  I have literally spent hours on it and got no where.
What I tried most recently was:
Wk 1:   Mtg 1    Mtg 2    Mtg 3    Mtg 4    Mtg 5    Mtg 6
        GATL1    GATL2    GATL3    GATL4    GBTL1    GBTL2
        GCTL4    GCTL3    GCTL2    GCTL1    GBTL4    GBTL3
        GDTL1    GDTL2    GDTL3    GDTL4    GETL1    GETL2

Wk 2:   Mtg 1    Mtg 2    Mtg 3    Mtg 4    Mtg 5    Mtg 6
        GETL2    GATL1    GATL2    GATL3    GATL4    GBTL1
        GCTL3    GCTL2    GCTL1    GBTL4    GBTL3    GBTL2
        GCTL4    GDTL1    GDTL2    GDTL3    GDTL4    GETL1

etc. etc. for 6 weeks.  The only problem is that this didn't ensure that everybody met with everybody else.
Very grateful for your assistance.

This is not a complete answer. I've renamed the people 1 through 18. The five groups are:
1 2 .. 3 4 5 6 .. 7 8 9 10 .. 11 12 13 14 .. 15 16 17 18
Here are the meetings in Week 1: 
1 3 7 11 ... 2 4 8 15 ... 5 9 12 16 ... 6 13 17 ... 10 14 18 
Week 2:
1 4 10 17 ... 2 5 7 14 ... 6 8 11 16 ... 3 12 15 ... 9 13 18 
Week 3:
1 5 13 15 ... 2 9 11 17 ... 3 8 14 ... 4 7 12 18 ... 6 10 16 
Week 4:
1 8 12 18 ... 2 3 13 16 ... 4 9 14 ... 5 10 11 17 ... 6 7 15 
Week 5:
1 6 9 ... 2 10 12 ... 
Week 6:
1 14 16 ... 2 6 18 ...
I don't know whether one can complete the Week 5 and Week 6 schedules to have everyone meet everyone. I know that 1 and 2 have met everyone, and most of the others have met 10 of the 14 they have to meet. 3 still has to meet 9, 10, 17 and 18; 4 has to meet 11, 13, 16; and so on. 
