I'm not sure if this is the right place to post the question.

How many ACTUAL chess board configurations are their?  I HAVE been able to find sites with answers to this question that include boards with extra pieces (like a board full of white Kings) (http://ioannis.virtualcomposer2000.com/math/EveryChess.html), but this is not what I am asking.  There can not be extra pieces.

It is OK, however, if one or more of the board configurations in the answer would be impossible to get to in a normal chess game.

So, to sum up:
- Total number of possible chess board configurations.
- Using at most 1 King, 1 Queen, 2 Rooks, 2 Knights, 2 Bishops, and 8 Pawns. (black and white).
- Some Impossible boards are OK (such as a board with no black king, or any board that could never happen in a normal game.  Even boards with overlapping pieces would be OK I guess, although not preferred).

BONUS: If you can exclude impossible boards in the answer, more power to you!

12

Various numbers get bandied about. 10¹²⁰ seems to crop up a lot.

As far as I can see, 13⁶⁴ (Fermat's Last Pizza) ought to be the upper bound (that's about  10⁷¹  ) - and that includes boards with way too many pieces.

This thread - http://www.chess.com/forum/view/general/how-many-different-chess-positions-are-there - discussed an approach for calculation based on enumerating sets of valid pieces (e.g. 2 pieces: 2 kings; 3 pieces: 2 kings + white pawn; 2 kings + black pawn; 2 kings + white bishop, etc.) all the way up to 32 pieces.

Ziryab wrote: If you google "chess position statistics," you should find the website of François Labelle at UC Berkeley who is working on computer modeling and mathematics and has an interest in chess. You'll see that he is carrying Claude Shannon's research much further.

According to Labelle, there are no more than 10⁵⁰ distinct board positions.

give each piece a name or a numver, you'll have 32 different pieces on the board.

you can put the first piece on  one of the 64 squares, the second in one of the remainng 63 squares etc

the result is

64*63*62*...33*32

=64!/31!

1.27e89/8e33= 1.58e55

but i'm not considering pawn promotion and as already told i'm counsidering each pawn different from another.

I had some fun with this on Chess.com:

"Is Chess Infinite?"

- Zug

You're also only considering using ALL 32 pieces. What about positions with fewer than 32 pieces?

you are right

so a better solution is sum of all combination with 32 pieces + 32 *sum of all combination with 31 pieces + 32^2 *sum of allcombination with 30 pieces etc.. +32^31 sumof all combination with 1 pieces

wow

64!/32! +32 *64!/33!+ 32^2*64/34!+.. +32^31*64!

= 64!/32!* (1+32/33+32^2/(33*34) + 32^3/(33*34*35)+....+ 32^31/(33*34*35*...*64))

=4.8e53*(1+0.97+0.91+0,83+0.74+0.65+0,57+...)

=(circa) 5e54

I hope

the number is far from perfect.. but it's near other approximation

I think I understand... but could you figure it out without considering each pawn different from each other (and the other pieces)?

i've seen it written in some chess books, and some chess blogs (the questions arise thus - "with all of the top players, isnt it just now a matter of memory?") ...that after about move 8, the positions grow so rapidly that by the middle game the possibilities outnumber the atoms in the universe

Zug's article is a very fun read, however some quick research shows 10^50 as a lower bound on the state-space complexity of chess, aka the number of legal positions.

The game tree complexity is indeed Shannon's number, 10^120 which is all possible positons.

Note that Go's game tree complexity is somewhere in the neighborhood of 10^360, which is fantastically larger, though this could possibly be due to the average Go game length being nearly twice that of a chess game. (80 vs. 150)

Way over my head .

You are on about permutations?

If so, one can not calculate untill a given place of stay!

But these numbers include boards with more pieces then there really are (like 64 Black Kings).

my math skills are good, but I'm not the best with combinatoric math.  :)

yes, but in all fairness there really are only 12 pieces, since all but king and queen have another.

Only 12 distinct pieces - but it's better to use the number 13 in calculations, so that empty squares are included.

but why? if the pawns could promote..

There are a maximum of 32 pieces on a chessboard, and a minimum of two pieces. When counting board positions, you also have to take into account that a square may be unoccupied.

Many board positions are duplicates, since the pieces are duplicated. In order to count DISTINCT positions, one uses the count of distinct pieces (i.e. 12 + empty). Promotions don't alter the number of distinct pieces. In mathematical terms, we would be counting combinations.

To count all board positions, including duplicates, one counts the pieces including duplicates (i.e. 32 + empty). In mathematical terms, we would be counting permutations.

I took out my wooden set and board and have started on the problem. will post the answer as soon as I finish.

PS. Admin- can u pls. extend my vacation time for me to complete this, thanks.

Under those stipulations the problem isn't all that hard. For all 32 pieces it is 63!/(32!*8!*8!)=4.63*10^42 unless I've made a mistake in the calculations. The rest will take a little longer to work out but I'm working on it. My number is lower than the others as I've took into account that all pieces aren't different.

PS: 63 is not a typo.