Hello!!

Great question!

I think you referred to the part 6C2 * 4C2 * 2C2 / 3! 

Here we go.

(A,B),(C,D),(E,F)

(A,B),(E,F),(C,D)

(C,D),(A,B),(E,F),

(C,D),(E,F),(A,B)

(E,F),(A,B),(C,D)

(E,F),(C,D),(A,B)

As you see from the list above, six people divided into groups of 3 in (A,B), (C,D), and (E,F) can be laid out in 3! ways. We don't want overcount, so we get rid of the overcount by dividing it by 3!. 

Hence, we only compute (A,B), (C,D), (E,F) just once, not all six ways!

Yay! Hope this helps!