# choosing two not empty groups out of $n$ people

How many ways are there to choose two not empty groups out of $$n$$ people? (each person is different) I thought I can maybe do this: $$\binom{n}{2}\cdot2\cdot2^{n-2}$$ First choose $$2$$ people to put in each group (and there are two options to put them) and the rest will have two options. What do you think?

• Are the groups required to be disjoint? Are the groups distinguishable? Jul 2, 2019 at 13:21
• Hint: there are three categories of people. Those in Group $A$, those in group $B$ and those in neither. ( I am assuming you wanted $A,B$ to be disjoint). If $A,B$ are not labeled then you have to divide by $2$ to account for the symmetry between them.
– lulu
Jul 2, 2019 at 13:22
• You have a lot of double counting. The first two people you choose may be in different groups when two other people are chosen first. Jul 2, 2019 at 13:22
• Is everyone suppose to be in one of the two groups? Your approach suggests that, and it's how I interpret the question, but others apparently do not. Jul 2, 2019 at 13:25
• As you can see from the comments, your question is unclear. Can you add more detail? It's always a good idea to work numerical examples. What do you think the answer is for $n=2$ or $n=3$? With my interpretation I see $2, 12$ respectively but of course you might have meant something else.
– lulu
Jul 2, 2019 at 13:26

$$(2^{n-1}-1)$$ if group labels don't matter, or $$(2^{n-1}-1)\times 2$$ if labels do matter.
Picking the leader for the first group, then picking the leader for the second group, then deciding how to split the remaining people in $$n(n-1)2^{n-2}$$ ways, or if labels on groups don't matter, in $$\binom{n}{2}\times 2^{n-2}$$