These are tricky. Hopefully I didn't make a mistake.

Yes. The answer is .

Once you've chosen your 20 people, you now have to count the number of ways to pair these people, which is . So it should be .2) Form 10 pairs from the group? (This is non-gender specific so you first choose 20 People from the group =C(30,20)? but what do you multiply it with to get the pairs? Or am I completely wrong about the first half?)

Recall that is the number of ways to choose objects from when the ordering doesn't matter. If there is more than one grouping step, multiply by the number of ways each step can be done.

Here, dividing the group into two means that if you have (A,B,C,D,E,F), choosing (A,B,C) is the same as choosing (D,E,F). So you have to divide by 2 to get rid of the duplicates.3) Divide the group into two groups (group 1 and 2) of equal size? (Is this just choosing 15 people from 30 = C(30,15)?)

So the number of ways to divide into two groups is .

The question is a bit confusing though, especially in light of the next question...

Perhaps what is meant is how many ways to make two groups of equal size? That is to say, you can have a group of 1 and 1, a group of 2 and 2, a group of 3 and 3, etc. In that case you would have

This doesn't make sense. If you have a group of 15 with 7 women, you will have 8 men and vice versa. So I don't see how each group can have as many men as women.4) Divide the group into two equal groups, where each group in its own has as many women as men in it? (would this be taking 7 men from 15 and multiplying it by taking 7 women from 15? I'm now officially lost and hopelessly confused?)

If you allow for groups of 2, 4, 6, etc., and you have a group with an even number in each group, you can have an equal number of men and women. So, following the last problem, you would have

Hmm. Only group 1 needs 4 men?5) Divide the group into two groups of equal size, such that group 1 contains at least 4 men? (I don't have a clue. Choose 15 from 30 people multiplied by C(15,4)?)

Well, we can take what we had before and modify it. First of all, clearly we can't have any group with less than 4 members. So the number of ways to create two equal groups with more than 4 members is:

Now we're in a bit of trouble because it does matter if they go into group 1 versus group 2. So we need to count the ways to put 4/15 men in group 1 and then we can choose the rest of the groups freely. The number of ways to choose 4 men to be in the group of size 6 is

because we have to choose 4 men from 15, then we can choose the remaining 2 from any of the 26 left to make up the rest of the group, and then we can choose any 6 of the remaining 24 for group 2. So we have

Now we don't have a size requirement on the groups.6) Divide the group into two groups each having size at least 1?(C(30,1)?)

So you can have a group of

So that gives us

I think these are right. But these kinds of problems are always really, really tricky so I could be wrong.