Find the number of possible arrangements in Fisher Random Chess. The diagram below is one possible arrangement.
In a legal arrangement, the White's position must satisfy the following criteria:
- Eight pawns must be in the $2^{nd}$ row. (The same as regular chess)
- Two bishops must be in opposite colored squares (e.g. $b1$ and $e1$ in the above diagram)
- King must locate between two rooks (e.g. in the diagram above, King is at $c1$ and two rooks are at $a1$ and $g1$)
The Black's position will be mirroring to the White's.
The answer is $\boxed{960}$. This is why the Fisher Random Chess is also called Chess 960.
The answer can be found using the multiplication principle.
First, let's place the two bishops. Each has 4 choices, hence there are $4\times 4=16$ ways. After having placed bishops, there are $6$ places left.
Secondly, let's place the two rooks and the king. By the rules, their relative positions must be R-K-R. Hence, we just need to select $3$ spots from the $6$ choices: $C_6^3 = 20$. This leaves $3$ places for the two knights and the queen.
Then, there are $3$ possible ways to put the queen.
Finally, there are two places for the two knights.
Hence, the final answer is $16\times 20\times 3=960$.