Permutation Problem - Seating with Empty Chairs There are 3 men and 3 women to be seated in a row of 10 chairs. In how many diﬀerent ways can they be seated if one man must be seated at each end of the row?
I began by calculating $_3P_2 = 6$ for the possible combinations for the end seats. The book gives the answer as 10,080, which I obtained by multiplying the first result by $_8P_4 = 1680$. I am unsure, however, about this second part of the problem, and I really only got there after trial-and-error, so I don't really understand why it works. 
My understanding of the permutation formula $_nP_k$ is that it is for selecting $k$ objects from $n$ objects when order matters. So, if I were selecting $4$ people to fill $4$ remaining seats, it would make sense to me, but I don't understand where the $4$ empty seats come into play. Shouldn't there be many more possible permutations when accounting for the different positions of the empty chairs?

Ignore "permutation" formulas and just do this directly via rule of product.  (It is afterall from the rule of product that we get the permutation formulas in the first place).


*

*Choose which man sits at the far left end (three options)

*Choose which man sits at the far right end (two remaining options)

*Choose which seat from those left available the remaining man sits (eight remaining options)

*Choose which seat from those left available the youngest woman sits (seven remaining options)

*Choose which seat from those left available the youngest remaining woman sits (six remaining options)

*Choose which seat from those left available the youngest remaining woman sits (five remaining options)


Multiplying the number of options for each step together yields the total:
$3\cdot 2\cdot 8\cdot 7\cdot 6\cdot 5 = 10080$
