How many group combinations of 3 can I make of 6 people? How many different combinations of groups of $3$ can I make of $6$ people: $A, B, C, D, E,$ and $F$?
{(group1), (group2), (group3)} 


*

*The order of each group doesn't matter

*The order of the whole set of groups doesn't matter either


One combination:
$(A,B), (C,D), (E,F)$
Another valid combination
$(A,B), (C,E), (D,F)$
The order of the 3 groups doesn't matter so:
This combination: $(A,B), (C,E), (D,F)$, is the same as: $(C,E), (D,F), (A,B)$
The order of each of the 3 groups consisting members doesn't matter too. 
This combination:  $(A,B), (C,E), (D,F)$, is the same as: $(B, A), (E, C), (F,D)$

What formula can be used to determine this?
 A: The solution to your problem is 
$$\frac{\binom{6}{2}\binom{4}{2}\binom{2}{2}}{3!}=15$$
Why does this work? Well, there are $\binom{6}{2}$ ways to pick the first pair of objects, $\binom{4}{2}$ ways to pick the second pair of objects, and $\binom{2}{2}$ ways to pick the third pair of objects. But since you don't care about the order in which you pick these pairs, and there are $3!$ possible orders, you must divide the product of these numbers by $3!$, and use
$$\frac{\binom{6}{2}\binom{4}{2}\binom{2}{2}}{3!}=15$$
A: The solution by @Frpzzd is perfectly fine. This is just another approach I read somewhere.
You first start by putting them all in a line and starting from the first,  selecting the two adjacent to form a group. This gives you a total of 
$$N=6!$$
Now since the ordering of groups don't matter, you divide by $3!$ to get
$$N=\frac{6!}{3!}$$
Also since the group members inside any group can be interchanged with each other, or swapped; so we have two options for each group - swap or don't. As there are $3$ groups
$$N=\frac{6!}{2^3\cdot 3!} = 15$$
A: Here is another way to find your answer. This method also directly implies a way to write a computer program to write out all the answers, in lexicographical order without any repetitions. This also is the shortest way to write the answer.
First, we consider the "first" person in the group. We can choose "first" in any way we choose--let's use the first one in alphabetical order by name. So we have chosen A. This person must be linked with another person. There are 5 other people so we have 5 choices here.
Now we choose the "next" person other than the two that we have already put into a group. If we chose B last time, that means C, otherwise that means B. We now choose a partner for that person, and there are 3 people left, so we have 3 choices.
Next we choose the "next" person other than the four that we have already put into a group, again using alphabetical order. Now there is only one other person, so that is the last partner we choose. This was 1 choice.
The number of choices was set, regardless of the actual choices we had made previously. Therefore we multiply those numbers, and the total number of possibilities was
$$5 \cdot 3 \cdot 1 = 15$$
Note that if we had $n$ people to break into pairs, then the number of possibilities is
$$(n-1) \cdot (n-3) \cdot (n-5) \cdots 3 \cdot 1$$
This is the double factorial or semifactorial of $n-1$, written as $(n-1)!!$. So the final, briefest answer to your question is
$$(6-1)!! = 5!! = 5 \cdot 3 \cdot 1 = 15$$
