How many group combinations of 3 can I make of 6 people?
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 $$\frac{\binom{6}{2}\binom{4}{2}\binom{2}{2}}{3!}=15$$