I'm not totally sure, but I'm interested to know the answer, so I'll give it my best shot. I think you would calculate the number of combinations by giving yourself 6 spaces (one for each position) and filling in each position with the total number of people who could possibly stand there. IF any of the mobsters could stand in any of the positions, your setup would look like this:
(6 mobsters) * (5 mobsters) * (4 mobsters) * (3) * (2) * (1)
Going from left to right, you subtract one for each slot, because, for instance, once you've filled the first slot, there are only 5 mobsters left who could possibly fill the 2nd slot, and once you fill the second slot, there are only 4 mobsters left who could possibly fill the third slot. To calculate the number of combinations, you multiply all of the numbers in these slots together-- 720.
BUT, the rules of your problem stipulate that Frankie must stand behind Joey, which means that Frankie cannot stand in the front, and Joey cannot stand in the rear. Therefore, for the first slot (we'll call it the rear), there are only 5 possibilities, so your setup would look like this (note that we include Frankie in the total number of possible mobsters only after we've filled the first slot--this is why the number 5 appears twice):
5*5*4*3*2*1
The total number of combinations here is 600. But this is where I get a little lost, because while I can account for Joey not being able to stand in the rear, I'm not sure how to account for Frankie not being able to stand in the front. You can't subtract 1 from the final slot, because this would yield zero. Perhaps accounting for this isn't necessary (perhaps subtracting 1 for the first slot accounts for both front and rear), or perhaps I've done all of this incorrectly.
