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?
Best Answer
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 $$\frac{\binom{6}{2}\binom{4}{2}\binom{2}{2}}{3!}=15$$