that led me to him being able to be in four spots (4!) but not in two spots (2!), out of six spots (numerator)
Unfortunately this isn't the right way to translate that situation into numbers. We can't simply divide by (the number of spots Joey can be in)! x (the number Joey can't be in)!. Think: this would mean that if Joey could be in all six spots, then we'd divide by 6! and there would be only one possibility - nonsense! I think you're mixing this up with something else: if we order objects and some of the objects are the same, then we divide by the factorial of that number. For more on this check out chapter 4 of the Number Properties Strategy Guide.