# Need help with 2 combinations of 3

I have a real-world math problem that may be simple for some of you but we are counselors and cannot figure it out. I have a group of 6 people who need to divide into 2 groups of 3 each week. Over the weeks the groups need to change so that the small groups cover each possible combination of three over time (everyone gets the opportunity to meet with others in small groups to get the most variety of interaction). This goes on indefinitely so there would be a cycle that repeats when it is complete.

Does anyone know how we can set this up to easily get the combinations in a grid or something? Thank you!

The easiest way to generate all the possible groups of three is to start with ABC (assuming our people are ABCDEF).

We then keep the first two the same and increase the last one and do this as much as possible to get ABD, ABE, ABF.

At this point we increase the second value and reset the third giving us ACD. Note that for this method we always keep the "people" in alphabetical order. This is how we ensure uniqueness since ACB has already been covered with ABC. You can think of this as a bit like counting where you keep increasing the right most item until you can't any more and then increment the one to its left.

This will give you: ABC, ABD, ABE, ABF, ACD, ACE, ACF, ADE, ADF, AEF, BCD, BCE, BCF, BDE, BDF, BEF, CDE, CDF, CEF, DEF.

Now the important thing to note is that these are the unique groups but of course you have two groups a week. The first and last are matching pairs, as are second and second last, etc. So you just need to loop through the first ten and the remaining ten groups will be their compliment.

• This was very helpful - I was able to plug in and replace our names for the letters and I think we got it. thanks Aug 20, 2013 at 19:53

When choosing 2 groups of 3 from 6 people, it is sufficient to look at how many different ways we can choose group 1, since group 2 is implicitly chosen by our choice of group 1. There are 20 possible groups of 3 that can be chosen. @Chris kindly pointed out, however, that of these 20 possible groups, they each pair up with its complementary pair each week, so if you rotated groups every week, you could go 10 weeks with completely different groups until you would need to repeat this again.

This result comes directly from the binomial coefficient (We just evaluate "6 choose 3"), and if more intuition is required, take a look at this

• You can only go for ten weeks since you have two groups each week. You have already said that group 1 defines group 2 but haven't taken that into account when figuring in how long before the rotation finishes. Unless each distinct group has to go in slot 1 and slot 2 in which case 20 weeks would be the right period but I don't think the question suggests that. Aug 20, 2013 at 18:53
• Yea, I just realized that....Thanks for pointing it out =) Aug 20, 2013 at 18:54

If the people are $ABCDEF$ call the group with $A$ in "Group 1". The other group is "Group 2" and consists of the remaining people.

There are five ways to choose the second person in Group 1, and then four ways to choose the third person - but this double-counts since choosing $BC$ is the same as choosing $CB$. So the number of ways of splitting into groups is $\cfrac {5\times 4}{2}=10$.