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!
 A: Call the first group A and the second group B.
We can start by assuming that no group can be empty or else the second group would not technically be a group. Thus the combinations
$A = 6$ $ B = 0$
$A = 0$ $B = 6$ 
cannot exist.
Forming 2 groups can be done by putting a certain number of people in the first group, and then automatically, we have formed the second group. That means that you only ever have to worry about calculating the number of ways to do so for 1 group and that will take care of the second group.
Now to form the first group you can:


*

*Take 1 person out of 6 and place him in A. B then has 5 people. There are $\binom{6}{1}$ ways of doing this.

*Take 2 people out of 6 and place them in A. B then has 4 people. There are $\binom{6}{2}$ ways of doing this.

*Take 3 people out of 6 and place them in A. B then has 3 people. There are $\binom{6}{3}$ ways of doing this.

*Take 4 people out of 6 and place them in A. B then has 2 people. There are $\binom{6}{4}$ ways of doing this.

*Take 5 people out of 6 and place them in A. B then has 1 person. There are $\binom{6}{5}$ ways of doing this.


We're almost done. Notice 2 things however.
First of all, the grouping of the second group is implicit in the calculation. In fact, taking the first example, if you place 1 person in A there are $\binom{6}{1}$ of doing this, and then $\binom{5}{5}$ ways of putting the remaining 5 people into B. Since $\binom{5}{5} = 1$, each binomial coefficient you see is actually a product of 2 binomial coefficients where one is equal to 1. 
Second of all, and most importantly, the order of the teams themselves does not matter. If we choose, say, the pair:
$\{A = 5,B = 1\}$
that's exactly the same pair as this:
$\{A = 1,B=5\}$
Because the order of the teams themselves does not matter, we must divide by 2! = 2, the number of different orders we can put the 2 teams in, because some of the different orders are in fact the same set of teams. When you just multiply your binomial coefficients together (the explicit and implicit one), however, these all get counted as distinct. Dividing by $2!$ collapses these all into a single arrangement.
That means that by the sum rule, there are:
$$
\frac{\binom{6}{1} + \binom{6}{2}+ \binom{6}{3}+ \binom{6}{4} + \binom{6}{5}}{2} = \frac{62}{2} = 31
$$
ways of splitting 6 people into 2 groups.
A: 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.
A: 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. 
