Given some electron configuration, I know the following formula can be used to determine the number of microstates:
$$\text{# mircostates} = \frac{(\text{# electron positions})!}{((\text{# electrons})!(\text{# positions} - \text{# electrons})!)}$$
So for 2 electrons in a $\mathrm{p}$ orbital it would be $6!/(2!4!) = 15$, or for two electrons in a $\mathrm{d}$ orbital it would be $10!/(2!8!) = 45$.
I'm a little confused how to apply this for a situation where all the electrons are not in the same $l$ level.
For a configuration like one electron in $\mathrm{s}$ and one in $\mathrm{d}$ would the number of microstates be the product of the above formula for each $l$ level?
$$\frac{2!}{1!1!}\cdot \frac{10!}{1!9!} = 20$$
Or for one electron in $\mathrm{s}$ and one in $\mathrm{f}$:
$$\frac{2!}{1!1!}\cdot \frac{14!}{1!13!} = 28$$
I assumed it would work like this because the formula reminds me of multiplicity, so it would make sense for it to be a product. But it also occurred to me that the 2 $l$ levels could be combined for the formula, so for a $\mathrm{(s^1)(d^1)}$ electron configuration, the number of positions would be $$\frac{(10+2)!}{2!(10+2-2)!} = \frac{12!}{2!10!}= 66,$$ which is must bigger than what I had before.
Can anyone clear up how to apply this formula to a case like this?
For clarification, the exact specification in the problem was:
Imagine a $\mathrm{s^1\, f^1}$ electron system (that is, 1 s electron and 1 f electron; no p or d electrons). How many total microstates exist for this system?
I think this means that the one $\mathrm{s}$ and one $\mathrm{f}$ electrons are each confined to their $l$ level, which is why I thought the multiplicity of microstates would be the product of the above formula for each, but I'm not sure.