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?

First you choose $2$ man out of $3$ which are sited at the each end and multiply by $2$. Now from the other $8$ chairs you choose $4$ chair and then arrange the rest $4$ people on this chairs (take all permutation of these $4$). So the answer is: $${3\choose 2}\cdot 2\cdot {8\choose 4}\cdot 4! = 10080$$ 
