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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.

Alex
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    Knights on different colours? I take it you don't play chess? – bof Jan 01 '15 at 21:56
  • Hint: if we know which three squares are occupied by the king and rooks, then we know which of those squares the king is on. – bof Jan 01 '15 at 21:59
  • this problem isn't suppose to make any sense from the chess point of view. Only combinatorics. I'm not sure I understand your second comment. – Alex Jan 01 '15 at 22:01
  • The question (with bishops not knights required to be on squares of different colour) does make sense from the chess point of view: it's asking for the number of possible starting positions in Fischer Random Chess aka Chess 960. What is it that you don't understand about my second comment? – bof Jan 01 '15 at 22:06
  • I don't get it: $\binom{6}{3}$ is the number of ways to chose 3 distinct positions out of 6 such that the order of the positions within the choice doesn't matter. How come it doesn't matter when we want the King to be strictly between rooks? – Alex Jan 01 '15 at 22:13

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You choose $3$ squares out of the $6$ that are not occupied by the knights. Then the king has to be on the middle of those three squares.

Arthur
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  • why? no, really, why does it exclude the k-r-r arrangement? – Alex Jan 01 '15 at 22:28
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    I meant that after you pick $3$ squares there is exactly one way to put the king and rooks on those squares according to the rules of the problem. Namely with the king on the middle square. – Arthur Jan 01 '15 at 22:37
  • So if we wanted an arrangement that excludes king in the middle, we'd have $2 \binom{6}{3}$ choices? – Alex Jan 01 '15 at 23:58