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!
Let's calculate the probability, then convert that to odds. 

On a fair die, half the numbers are even and half the numbers are odd. So, the probability for a single roll of getting an even number or an odd number is $\dfrac{1}{2}$. 

The probability for a specific roll are unaffected by previous rolls, so we can apply the product principle and multiply probabilities for each roll. Each roll has probability of $\dfrac 1 2$ of obtaining the desired result. So, we have:

$$P(E,E,O) = \dfrac 1 2 \cdot \dfrac 1 2 \cdot \dfrac 1 2 = \dfrac 1 8$$

Now, the probability of that not happening is $$1-\dfrac 1 8 = \dfrac 7 8$$

So, the odds are 7:1 against the desired outcome.
All of the above or below :-} answers are correct that is 50:50 for each throw so your "formula" is

  n
(2 -1):1 against where n is the number of throws