Author: Andreas Stabel
Date: 07:25:52 01/13/00
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On January 12, 2000 at 18:34:49, Dan Ellwein wrote: >Hi > >Just wanted to bounce this off of the group and see if this is an accurate >representation of how many (non redundant) pawn positions there are in chess... > > >(0,8)(1,7)(2,6)(3,5)(4,4) 8P 48x47x46x45x44x43x42x41 x5 = _______ >(0,7)(1,6)(2,5)(3,4) 7P 48X47X46X45X44X43X42 x4 = _______ >(0,6)(1,5)(2,4)(3,3) 6p 48X47X46X45X44X43 x4 = _______ >(0,5)(1,4)(2,3) 5P 48x47x46x45x44 x3 = _______ >(0,4)(1,3)(2,2) 4P 48x47x46x45 x3 = _______ >(0,3)(1,2) 3P 48x47x46 x2 = _______ >(0,2)(1,1) 2P 48x47 x2 = _______ >(0,1) 1P 48 x1 = _______ > > >Number of non redundant pawn postions in chess -- TOTAL: _______ > >haven't done the math on it yet, but it looks like about 75 trillion... > >thanks... > >PilgrimDan I don't think it's that easy. My numbers for 8 white and 8 black pawns are: White pawns : 48 BNM 8 = 377348994 Black pawns : 40 BNM 8 = 76904685 (BNM is the binominal coeffisient) Multiplying these two numbers give 29019905518636890 This calculation has to be done for all possible combinations of white and black pawns (8,7 - 7,8 - 8,6 - 7,7 - 6,8 ...) and then added to get the total number. Regards Andreas Stabel
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