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.
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!)
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.
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!).
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?
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.
Brain Busting Variant!!!!!!!!!!!!!!!!!!!!!!
07 Feb '14 21:35
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