Trying to Split Six People into Two Groups and Have Each Person Meet

In summary, the conversation discusses the challenge of creating a virtual meeting with six people in two groups of three, where everyone must see each other at least once. Despite attempting various solutions, it is determined that after three rounds, it is impossible for everyone to meet. The group discusses different possibilities and concludes that there is no easy formula or calculation to solve the problem, but a computer program can help.
  • #1
The Head
144
2
Real World Application here. I'm creating a virtual meeting involving six people and will have the first round include two groups of three. Then we'll switch a few times. I tried hopelessly to do it such that after three rounds, everyone would see everyone else at least once. Was so close but always missed one or two pairs. It's not a straight 6C3 problem because if I have people A, B, C, D, E, F, I just care that A meets with D & F, but I don't necessarily need ADF to be a group. It could be ADE and then ABF.

Anyway, just trying to figure out how many rounds I'd need to make sure everyone interfaced with everyone else at least once, or if anyone has a suggestion that's better than guess and check.

Note, there may be a complication where only three of the six can "host" the meeting due to who has access to the full software, but honestly just want to figure this out first.
 
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  • #2
It's impossible after three rounds. Let's look at "A" first: They can meet two people per round and have to meet five people. Without loss of generality we can assume they meet B and C in the first round, D and E in the second, and F and someone else in the third round. This already fixes the first two rounds completely:
ABC DEF
ADE BCF
The third round has a group of "AF". So far B didn't see D and E, so we would need to make a group BDE, so our third round is ACF and BDE. But that means C doesn't see D and E.
 
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Likes The Head, PeroK and phinds
  • #3
mfb said:
It's impossible after three rounds. Let's look at "A" first: They can meet two people per round and have to meet five people. Without loss of generality we can assume they meet B and C in the first round, D and E in the second, and F and someone else in the third round. This already fixes the first two rounds completely:
ABC DEF
ADE BCF
The third round has a group of "AF". So far B didn't see D and E, so we would need to make a group BDE, so our third round is ACF and BDE. But that means C doesn't see D and E.

Thanks, that's helpful. I'll have to modify the format. I assume there's no formula or calculation I can do that's readily accessible that can help me check once I've adjusted?
 
  • #4
Not that I'm aware of. A computer program can do it, of course.
 

1. How many different ways can six people be split into two groups?

There are 15 different ways to split six people into two groups, assuming that the order of the groups does not matter.

2. Is there a way to ensure that each person meets someone from the other group?

Yes, there is a mathematical solution known as the "round robin" method which guarantees that each person will meet someone from the other group.

3. Can the groups be split evenly, with three people in each group?

No, it is not possible to split six people into two groups with an equal number of people in each group.

4. How does the number of people affect the number of possible group combinations?

The number of possible group combinations increases exponentially as the number of people increases. For example, with six people, there are 15 possible combinations, but with 10 people, there are 945 possible combinations.

5. Are there any other factors to consider when trying to split people into groups?

Yes, there are other factors to consider such as the size and composition of the groups, the purpose of the groupings, and the preferences or characteristics of the individuals being grouped.

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