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?

  • $\begingroup$ Are the groups required to be disjoint? Are the groups distinguishable? $\endgroup$ – paw88789 Jul 2 at 13:21
  • $\begingroup$ 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. $\endgroup$ – lulu Jul 2 at 13:22
  • $\begingroup$ 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. $\endgroup$ – saulspatz Jul 2 at 13:22
  • $\begingroup$ 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. $\endgroup$ – saulspatz Jul 2 at 13:25
  • 1
    $\begingroup$ 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. $\endgroup$ – lulu Jul 2 at 13:26

Without loss of generality, among your people there is a youngest person.

Choose who else from among all other available people is also in that youngest person's group, noting that it cannot be everyone (otherwise the other group would be empty).

If you care about labels on the groups as well, then afterwards, decide which label is used for the group containing the youngest person.

$(2^{n-1}-1)$ if group labels don't matter, or $(2^{n-1}-1)\times 2$ if labels do matter.

As an aside, your answer would have been correct to a different but related problem where each group has a designated "leader." The persons you picked in the first step are being treated differently than those arranged at later steps.

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 $\binom{n}{2}\times 2^{n-2}$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.