If a fair die is rolled 3 times, what are the odds of getting an even number on each of the first 2 rolls, and an odd number on the third roll? If a fair die is rolled 3 times, what are the odds of getting an even number on each of the first 2 rolls, and an odd number on the third roll?
I think the permutations formula is needed i.e. $n!/(n-r)!$ because order matters but I'm not sure if n is 3 or 6 and what would r be?
Any help would be much appreciated!

Assuming the die is just marked even and odd rather than with numbers, there are eight orderings.  They range from all odd to all even:  OOO, OOE, OEO, EOO, OEE, EOE, EEO, EEE.  In your formula:  
$$\frac{n!}{(n-r)!}$$
You are missing that you don't care about the orders of the duplicates.  So you need 
$$\frac{n!}{(n-r)!r!}$$
The $n$ is the total number of rolls, the $r$ is the number of either evens or odds (it's symmetric).  But to get the total number of orderings, you need to add these:  
$$\sum\limits_{r=0}^n\frac{n!}{(n-r)!r!}$$
Now substitute 3 for $n$.  
$$\sum\limits_{r=0}^3\frac{3!}{(3-r)!r!}$$
Unrolling that, we get 
$$\frac{3!}{3!0!} + \frac{3!}{2!1!} + \frac{3!}{1!2!} + \frac{3!}{0!3!}$$
or 
$$1 + 3 + 3 + 1$$
So we have one all odds, three with one even, three with two evens, and one all even.  That's eight total.  
Another way of thinking of this is that there is only one ordering of all odd numbers or all even numbers while there are three places where the lone odd or even number can be.  
Of those eight, how many fit your parameters?  Exactly one, OOE.  So one in eight or $\frac{1}{8}$.  
As others have already noted, you could get that much more easily by simply figuring that you have a one in two chance of getting the result you need for each roll.  There's three rolls, so $(\frac{1}{2})^3 = \frac{1}{8}$.  
If you want to treat 1, 3, and 5 as different values and 2, 4, and 6 as different values, you can.  But it is much easier to think of them as just odd or even.  Because you don't want to try write this out for $6^3 = 216$ orderings.  And in the end, you will get the same basic result.  You will have twenty-seven OOE orderings, which is again one eighth of the total.  This is because there are three possible values for each, 1, 3, and 5 for the two odds and 2, 4, and 6 for the even.  And $\frac{27}{216} = \frac{1}{8}$.  
Permutations leads you down a harder path.  It's easier to think just in terms of probability or even ordering.  
