Chess Piece Combinations

Solution 1:

Assuming that the king and rooks do not have to be on consecutive squares, the question is about the number of legal arrays in Fischer Random Chess aka Chess960. The answer is (surprise!) $960$. Here is an explanation from the Wikipedia page:

Each bishop can take one of four positions, the queen one of six, and the two knights can assume five or four possible positions respectively. This leaves three open squares which the king and rooks must occupy according to setup stipulations, without choice. This means there are 4×4×6×5×4 = 1920 possible starting positions if the two knights were different in some way. However, the two knights are indistinguishable during play (if swapped, there would be no difference), so the number of distinguishable possible positions is half of 1920, or 1920/2 = 960. (Half of the 960 are left-right mirror images of the other half, however Chess960 castling rules preserve left-right asymmetry in play.)

Of these $960$ positions, just $108$ satisfy the further condition that the rooks be right next to the king. Namely, there are $6$ ways to place the RKR combo. Any placement of the RKR combo leaves three squares of one color and two squares of the other color, so there are $3\times2$ ways to place the bishops. The queen then has a choice of $3$ squares, and the knights take what's left; $6\times3\times2\times3=108$.

The requirement that the king be placed between the rooks is natural, because it permits a kind of generalized castling with an effect somewhat similar to castling in orthodox chess. Requiring the rooks to be placed right next to the king has no apparent chessic motivation.