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I need to know how many possible positions are there for 32 chess pieces to fit onto an 8x8 chess board including type of piece, and number of pieces, and color

Please show equation used

2006-09-16 09:35:00 · 7 answers · asked by psandels 1 in Science & Mathematics Mathematics

7 answers

It is about 10^120

2006-09-16 09:50:24 · answer #1 · answered by Barkley Hound 7 · 0 0

64! / (8!^2 * 2!^6 * 1^2 * 1^2)

(2*8 pawns, 2*2 rooks, knights, bishops, 2*1 king, queen)

The numerator is equal to 1 * 2 * ... * 63 * 64.
The denominator is
(8 * 7 * 6 * 5 * 4 * 3)^2 * 2^8, or
2^20 * 3^4 * 5 * 7.

Using the factors in den. to cancel those in the numerator we end up with

31 * 59 * 29 * 19 * 7 * 55 * 27 * 53 * 13 * 25 * 6 * 47 * 46 * 45 * 44 * ...

which is

1082680410561750 * 47!

or a little more than 2.8 * 10^74.

2006-09-16 11:11:55 · answer #2 · answered by dutch_prof 4 · 1 0

The simple upperbound is the permutations.

n select k for n=64 and k=32 is
factorial(n) over factorial(n-k)
or about
4.8222*10^53

The problem is they start from a single position, and have limited moves. That radically decreases the number of possible combinations.

Note: a single ordered list of pieces is sorted through all combinations of pieces using this logic. Making it square-assignment based makes it simpler.

2006-09-16 09:41:24 · answer #3 · answered by Curly 6 · 0 0

It has been solve many years ago by a "cray computer" and the number is so large that it is more than the observable atoms in the universe. Exact number is not worth knowing.

2006-09-16 17:09:55 · answer #4 · answered by Liwayway 3 · 0 0

In chinese chess, the cannon- because the chinese call for cannon is... pow! For western chess, i assume the knight, like the pow he can hop over different products, and he feels like a horse. particular chess rule- pawn climbs into the rook, then the rook can bypass because it likes, and the pawn receives to leap out on the top of a rook bypass if it needs. Hurray!

2016-11-27 19:12:47 · answer #5 · answered by ? 4 · 0 0

I don't know an equation for this (or even if there is one), but the number of possible positions is somewhere in the billions.

2006-09-16 09:41:18 · answer #6 · answered by Anonymous · 0 0

i think it will be 64C32

2006-09-16 09:50:33 · answer #7 · answered by raj 7 · 0 0

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