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A chessboard has 64 squares

  • 16 pawns
  • 4 rooks , bishops and knights each
  • 2 queen and king each

Total pieces are 32. Total squares are 64. (No pieces are captured) Let's now focus on one piece One piece can fit on 64 squares The other can fit on the remaining 63 squares

... The last (32th) piece would be on 32 squares

So the answer would be 64!/32! Please confirm

PS : I am a newbie into these things so I don't really understand factorials really well... The positions can BE ILLEGAL .

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No, it wouldn't. You will have over counted, as many of the pieces are interchangeable (any white pawn is the same as any other white pawn). The denominator should have an extra factors of $2^6 \cdot (8!)^2$ to account for the interchangeable pieces.

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  • $\begingroup$ How did you calculate that ? Please tell $\endgroup$ Feb 10 at 5:28
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    $\begingroup$ There are six pairs of identical pieces (white/black, knight/bishop/rook), hence the $2^6$ factor. Then, there are two groups (black/white again) of eight interchangeable pawns, hence, $(8!)^2$. $\endgroup$ Feb 10 at 5:29
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    $\begingroup$ This is also done under the assumption that all 32 pieces are on the board. It will be different if the pieces need not be on the board. $\endgroup$ Feb 10 at 5:31

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