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If you play a standard chess game on a normal $8 \cdot 8$ chess board with the usual rules: How many different "board representations" can exist?

Upper bound: Well, you have 16+16 = 32 chess pieces and 64 fields, so $\frac{64!}{32!} \approx 4.8 \cdot 10^{53}$ is an upper bound. This would mean that all chess pieces are distinguishable (e.g. white knights are not distinguishable) and you can distribut them freely accross the field (white pawns can never be in the first row).

Lower bound: If you look only at the pawns, you can imagine them as a number in base 3 (they can not move, move one in front or move two in front). You have 16 pawns, so $3^{17} \approx 1.3 \cdot 10^8$ is a lower bound.

I would also be happy with the number of different game situations (which is higher, as you might care of the possibility of castling, promotion or capture).

I guess the explanation will not be easy, so please provide a source.

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We will probably never know the exact number, even by computer search - I think the only fully reliable way to demonstrate that a position is legal is by proof game. There are positions in which it is tricky enough to determine legality that there is a type of puzzle based on this idea, frequently called a retrograde analysis study. janko.at/Retros/index.htm has some examples. en.wikipedia.org/wiki/Shannon_number provides some estimates. –  David Spencer Jul 12 '12 at 11:31
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Oh, and although your upper bound is probably above the actual number, it doesn't seem to take into account promotions or captures. –  David Spencer Jul 12 '12 at 11:43

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