Ok so my Stats professor never explains how to do this stuff

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The question is really asking: how many six-number combinations are there? (Order doesn't matter, and a number cannot appear twice.) Well, then we're asking: how many six-element subsets are there of a 27-element set? We've seen this type of question before.

The answer is: there are
27 choose 6 = (27!) / (6! * 21!) = 296010.

Well, then any one of these combinations has a probability of 1 / 296010 = 3.3782... * 10^(-6).

Note: I actually don't know anything about lotteries, so I'm just assuming that the order of the numbers doesn't matter. If the order matters, then the answer is different.

What we are asking here is: how many ways can you put 6 objects in order? The answer is 6! = 720.

This is a special case of a more general principle. The general rule is: if you want to know how many ways you can put n objects in order, the answer is always (n!).

Well, how many ways can we arrange 8 objects into an order? It's the same situation as the last question, and the answer is 8! = 40320.

So the probability that a particular one of these orders happens is 1 / 40320 = 2.4801... * 10^(-5).

I can't answer this question unless I know whether "and" or "or" fits in the gap.

Let's recap. We know that the number of r-element subsets of an n-element set is always
n choose r = (n!) / (r! * (n-r)!)

Well, we can also write nCr instead of "n choose r". It's just a matter of notation.

Here's a new definition:
nPr = (n!) / ( (n - r)! )

It's not so easy to describe what nPr measures without using technical words. Here's an attempt:

If you have an n-element set named S, then nPr tells you the number of orderings of r elements of S, where an element cannot appear twice in an ordering.

But we don't need to know what nPr "means" in order to do this question. We just need to use the definition. By the definition:

10P3 = (10!) / (7!) = 720.

Thanks again for taking the time to go over this with me. I know a lot more know than I did before. =)