# I have N different objects and I choose g out of them

1. Aug 8, 2009

### qasdc

Hi!
I have the following questions and I would like some help.

I have N different objects and I choose g out of them (without repetition), avoiding permutations of the objects. This can be done in N!/(N-g)!g! ways.
I create another set of g objects in the same way.

So, I have two such sets of g objects and the question is what is the probability that these two sets have at least one object in common? What is the probability of having exactly one object in common?

My actual problem is that i don't count correctly the different ways that I can create two sets with one object i common.

2. Aug 8, 2009

### mXSCNT

Re: Combinations

Well, the probability that the two sets share at least one element is 1 minus the probability that the sets do not share any elements between them. The number of ways to choose 2 sets of size g each from a set of N objects such that the sets have no intersection can be counted by the following method:
1. Select a set of size 2g from the set of N objects
2. Select a set of size g from the set of size 2g, thus partitioning the 2g elements into two disjoint sets of size g

That should be enough hints to get you going on the first part. For the second part, count the sets that share exactly one element this way:
1. Select one element from the set of size N, to be the shared element
2. Select two disjoint sets each of size g-1 from the set of size N-1, using the method of the first part

3. Aug 9, 2009

### qasdc

Re: Combinations

Ok, thnx!

I now understand what was wrong the way I counted.