This is also a basic combinatorics question, but I don't understand part of its solution. We have a King, a Queen, 2 rooks, 2 bishops and 2 knights (each of the last three pieces are identical, i.e. both bishops are of the same color, etc).
In how many ways can we arrange them in one line of the chessboard so that knights are on squares of different colours and the King is between two rooks. The rest is 'don't care'.
Knights are easy - $4 \cdot 4$. I'm confused about the rook/king/rook part. Why is it $\binom{6}{3}$? How does it account for the fact that the King is strictly $between$ the rooks? If I needed to allocate the pieces rook/rook/king or king/rook/rook, would it be $2 \binom{6}{3}$ then? If I had three rooks instead, would it still be $\binom{6}{3}$?
The rest is of course $3 \cdot 1$.
I realize the question is quite basic, but I seriously can't make my head around this rook/king/rook allocation.