In how many ways can you split six persons in two groups? In how many ways can you split six persons in two groups?

I think that I should use the binomial coefficient to calculate this but I dont know how.

If the two groups has to have equal size, then each group should have three persons. Then the calculation would be: 

$${{6}\choose{3}}/2 = 10$$ 

But since the groups now can have different sizes I dont really know. I would guess that you have to add up multiple binomial coefficients, something like this:

$$ {{6}\choose{1}} + {{6}\choose{2}} + {{6}\choose{3}}=41$$

But the answer should be 31 so that does not seems to be quite right. I would like some suggestions on this!
I assume the two groups must be non-empty!  It helps to know that Alicia is one of the people. 

Joining Alicia will be any subset of the remaining people, except for the full set. So there are $2^5-1$ ways to choose the people who will be in Alicia's group.
Total number of combinations of 3 from 6 = 6C3 = 20
But this answer needs to be divided by two as it asks for number of groups so each Combo of three cancels out its opposite pair. 20/2 = 10.

Number of uneven groups = number of 1 and 5s and number of 2 and 4s
= 5C1 + 4C2 = 21.  Again splits of 5 and 1 and 4 and 2 are not considered as they are simply mirrors of the 1-5, 2-4 combinations.