1. ## Easy combination problem

Suppose there are six people and I want to divide them into groups of 3 and 3. How many different ways can this occur (order of the people does not matter).

I know its (6C3 times 3C3)/2!, but I am uncertain why that is over 2!. What general reasoning am I missing.

Thanks.

2. Why are you dividing by "2!"?

3. I don't know! My teacher said it. I thought I understood the concept, but that threw me off...

4. I think you might be confusing the general form of nCr with what you are doing here. Remember that nCr is really just: $\displaystyle \frac{n!}{r!(n-r)!}$, where n!/(n-r)! is the number of ways you can arrange "r" objects out of a total possible "n" objects, and dividing by r! allows us to cancel out all the "duplicates" which are only different arrangements of the same group of objects (this is where the idea of "order doesn't matter" comes from).

In your situation you are told that you have 6 objects, and want to make two groups out of them. Right off the bat you know that the total possible ways you can make these groups will be # of Group A multiplied by # of Group B. Right off the bat, you should recognize that if you are making even groups, no matter what combination of people in Group A will be, there will only ever be one way you can arrange the people into Group B (everyone left over goes into group B).

Total ways of arranging nCr into groups:

Group A - $\displaystyle \frac{6!}{3!(6-3!)}$

Group B - $\displaystyle \frac{3!}{3!(3-3)!}$

Total ways to arrange the two groups:

$\displaystyle (\frac{6!}{3!(6-3!)})(\frac{3!}{3!(3-3)!}) = 20$

The teacher (I hope), knows what they are talking about - however you also need to make sure that you know only understand the material but that you KNOW the material. Or else it's not going to stick. Memorization is cool for social sciences - but here, without truly knowing a bit of material - its just going to slip right out of your brain the first chance it gets.

5. Originally Posted by nankor
Suppose there are six people and I want to divide them into groups of 3 and 3. How many different ways can this occur (order of the people does not matter). I know its (6C3 times 3C3)/2!, but I am uncertain why that is over 2!. What general reasoning am I missing.
You divide by two because this is an unordered partition.
If you divided into a red group and a blue group then that is $\displaystyle \frac{6!}{(3!)^2}$.
That is the member of ways to arrange the string $\displaystyle RRRBBB$.

But in this case we do not have labeled groups, so divide by two $\displaystyle \frac{6!}{2(3!)^2}$.