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In how many ways can we place two opposite kings on an empty chessboard in accordance with the rules of chess?

I have drawn an diagram presenting two scenarios - when the first king takes away 4 squares (label A) and some scenarios when it takes 6 squares (label B). However, the squares taken away started to overlap. How would you solve this problem? enter image description here

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You are on the right track.

  • If the white king is in a corner ($4$ possibilities), the black king can be at any of $64-4$ positions.
  • If the white king is at the boundary, but not in a corner ($24$ possibilities), the black king can be at any of $64-6$ positions
  • In all other cases for the white king ($36$ possibilities), the black king can be at any of $64-9$ positions.

In total: $$4\cdot 60+24\cdot 58+36\cdot 55. $$


Alternatively: There are $64\cdot 63$ ways to place the kings without respecting the rules (except that they are on different squares). Subtract invalid positions: There are $7\cdot 8$ positions with the black king directly to the east of the white king; the same holds for west, north, and south. And there are $7\cdot 7$ positins with the black king south-east of the white king; the same for south-west, north-west, north-est.

Hence: $${64\cdot 63}-4\cdot 56-4\cdot 49. $$

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