In how many ways can a group of six people be divided into: 2 equal groups? 2 unequal groups, if there must be at least one person in each group? In how many ways can a group of six people be divided into:

a) two equal groups

I have $^6C_3 \times \space ^3C_3 = 20$
So, to choose the first group I have $6$ possibilities of which I am choosing $3$. For the second group, I have $3$ remaining people of which $3$ must be chosen -> hence $^6C_3 \times ^3C_3 = 20$.
But the answer is $\frac{^6C_3}{2}$ but I don't understand why you divide by $2$.

b) two unequal groups, if there must be at least one person in each
group?

Applying the same logic as before, I got:
$$(^6C_2 \times ^4C_4) + (^6C_1 \times ^6C_5) = 51$$
But the answer is $^6C_1 + \space ^6C_2 = 21$
Could anyone explain how to solve these/the intuition behind it? Thanks in advance!
 A: *

*If you're dividing 6 people into two groups $x$ and $y$, then $^6C_3$ would give you the total number of pairs of $(x,y)$ as well as $(y,x)$. To avoid the repetition of the pair $(y,x)$, you divide by two.

*The same logic applies here: when you split into groups, the remaining people whom you didn't select to form a group form the second group. To avoid repetition, the number of groups is $^6C_1 + \space ^6C_2 = 21$.

If you're still confused, the best way to grasp this concept is to take 4 people $a,b,c,d$ and see in how many ways you can split them into groups of 2 manually (by writing down all cases).
A: For the first question: your attempt is related to the situation where the two groups are labeled. So, if the goal is to divide people equally into group A and B, then the answer is ${6 \choose 3}$.
However, this question (seeing from the solution) does not presupposes that these group is labeled. In this case, having people 1,2,3 in group A and people 4,5,6 in group B is equivalent to 4,5,6 in A and 1,2,3 in B, for they divide these people in the same fashion. Because of this type of duality, you have to divide your answer by 2, the factorial of the cardinal of each cosets of this equivalent relation.
