Question: My maths is very rusty. I am looking to calculate how many unique combinations I can draw from a given set of names.

I understand the formula $\frac{n!}{(n-r)!(r)!}$ will give me the number of unique combinations of $r$ elements from the set with $n$ elements, where all combinations will contain at least $1$ different element.

What if I wanted to specify the minimum number of elements that had to be different?

For example, the number of combinations of $6$ names I could draw from $36$ names = $\frac{36!}{30!\,6!}=1,947,792$

Every combination would have at least 1 name different to every other combination. How do I work out the number of combinations if I wanted at least 2 or 3 names in every combination to be different.

Names in the set are unique and can't be repeated.

Thank you.

Answer: 
Unfortunately, the maximum number of combinations under such a restriction is not known in any closed-form manner. The topic of constructing combinations like this is the topic of Block Design. There is a theoretical maximum, but there's no guarantee that the maximum can be achieved for any particular number of chosen elements or number of elements available.
The idea of the [Hamming distance](http://en.wikipedia.org/wiki/Hamming_distance) between each of your sets of six names would be useful. Each of your sets of six names would be considered a string of length $n=6$ from an alphabet set with $q=36$ elements.

By specifying a minimum number of elements to be different, you are specifying the minimum Hamming distance $d$ between each of the strings. There is then an upper bound to the number of possible strings that are at least $d$ apart from each other, given by the [Hamming Bound](http://en.wikipedia.org/wiki/Hamming_bound), but it is not always achievable, as mentioned by Glen.