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1.  ketchuplover
G.O.A.T.
07 Feb '14 21:35
CHESS 1,625,702,400 aka Random Pawn Placement
2.  Ponderable
chemist
08 Feb '14 10:34
Originally posted by ketchuplover
CHESS 1,625,702,400 aka Random Pawn Placement
would you elaborate? I for one can't do anything with the information provided.
3.  dzirilli
Duchampion
08 Feb '14 14:42
I'm assuming that's the number of possible starting positions if you place the pawns randomly
4.  ChessPraxis
Cowboy From Hell
08 Feb '14 15:52
Originally posted by ketchuplover
CHESS 1,625,702,400 aka Random Pawn Placement
I saw this being discussed at another site.
5. 08 Feb '14 16:07
Originally posted by dzirilli
I'm assuming that's the number of possible starting positions if you place the pawns randomly
It does not seem to be. The number is equal to 8! squared, which can't possibly be the number of ways to place pawns on the board. It could be a miscalculated number of ways to place all the non-pawns on their respective starting ranks -- there the mistake would at least make sense.
6.  wolfgang59
Infidel
08 Feb '14 21:40 / 1 edit
Originally posted by WanderingKing
It does not seem to be. The number is equal to 8! squared, which can't possibly be the number of ways to place pawns on the board. It could be a miscalculated number of ways to place all the non-pawns on their respective starting ranks -- there the mistake would at least make sense.
8! would be the number of ways of setting up your own pawns at the start
(given that the pawns are distinguishable)
Take into account the opponents permutations as well and you get (8!)x(8!)
7.  ketchuplover
G.O.A.T.
08 Feb '14 23:58
Originally posted by wolfgang59
8! would be the number of ways of setting up your own pawns at the start
(given that the pawns are distinguishable)
Take into account the opponents permutations as well and you get (8!)x(8!)
Does 8! = 1x2x3x4x5x6x7x8 ? Someone on another site thinks I am including missing pawns. Never thought of them.
8.  wolfgang59
Infidel
09 Feb '14 01:14
Originally posted by ketchuplover
Does 8! = 1x2x3x4x5x6x7x8 ?
YES!
9.  DeepThought
09 Feb '14 06:04
Originally posted by wolfgang59
8! would be the number of ways of setting up your own pawns at the start
(given that the pawns are distinguishable)
Take into account the opponents permutations as well and you get (8!)x(8!)
Except that pawns are indistinguishable so the correct number of ways of setting up the pawns at the start is 1.

If you had the pawns go anywhere where there aren't pieces, and avoiding the initial position involving a check then there are C(46,8)*C(38,8) ways of placing the pawns (C(N,M) = binomial coefficient = N!/(N-M)!M!).
10.  wolfgang59
Infidel
09 Feb '14 06:48
Originally posted by DeepThought
Except that pawns are indistinguishable so the correct number of ways of setting up the pawns at the start is 1.

Each pawn has its own character. I have one little fellow who is very aggressive ... always a tough call to start him on c2 or e2
11. 09 Feb '14 08:08
Originally posted by DeepThought
Except that pawns are indistinguishable so the correct number of ways of setting up the pawns at the start is 1.
Indeed. (I'm not checking the other half -- I have a hangover to cure.)
12.  wolfgang59
Infidel
09 Feb '14 08:40
Originally posted by WanderingKing
Indeed. (I'm not checking the other half -- I have a hangover to cure.)
Greasy bacon sandwich washed down with
a pint of fresh OJ & tonic water (50/50)

trust me
13.  sonhouse
Fast and Curious
09 Feb '14 13:09
Originally posted by DeepThought
Except that pawns are indistinguishable so the correct number of ways of setting up the pawns at the start is 1.

If you had the pawns go anywhere where there aren't pieces, and avoiding the initial position involving a check then there are C(46,8)*C(38,8) ways of placing the pawns (C(N,M) = binomial coefficient = N!/(N-M)!M!).
What does the 46,8 and 38,8 stand for? What does that number come out to?
I assume N=46? or 38?
14. 10 Feb '14 08:47 / 1 edit
Originally posted by sonhouse
What does the 46,8 and 38,8 stand for? What does that number come out to?
I assume N=46? or 38?
I think:

C(46,8)*C(38,8) ways of placing the pawns (C(N,M) = binomial coefficient = N!/(N-M)!M!).

translates into:

46!/(46-8)!8! * 38!/(38-8)!8!

So you apply the formula twice, once with N=46, M=8 and once with N=38, M=8. And then multiply those two.
Whether the outcome is correct, is another story.
15.  Ponderable
chemist
10 Feb '14 11:45
Since I am quite dumb:

You propose to put pawns on the second and seventh row rexpectively. Then you go and permuatet their position.

What is the difference to the nomrla person with the transmutations?