Author: José de Jesús García Ruvalcaba
Date: 14:02:21 08/09/99
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On August 09, 1999 at 15:29:44, blass uri wrote: > >On August 09, 1999 at 13:46:37, José de Jesús García Ruvalcaba wrote: > >> I finally sat down and make all the computations from scratch and then >>rechecked the results. >>Number of 5+1 tablebases: 40 >>Number of 4+2 tablebases: 200 >>Number of 3+3 tablebases: 120 >>Total number of six men tablebases: 360 >> If there is interest I can explain how I got the results, they are simple >>combinatorics problems. >>José. >Your calculation is wrong. >For example the first number is 70 > >if I represent P=1,N=2,B=3,R=4,Q=5 then the I can represent every 5+1 material >configuration(4+0 without kings) as 4 numbers and the number of possible >configuration is 70 > You are right. I forgot to add the tables with two equal pieces and the other two different (there are exactly 30 of them). There can be more mistakes, of course. Now that I use your techniques to count the 4+2 tables, I get 175 (it is five times the number of 4+1 tables), I wonder where I was wrong. My new numbers are: 5+1: 70 4+2: 175 3+3: 120 Total: 365. >I can represent pppp configuration by 1111 >pppn by 1112 pppb by 1113 > >You can count non decreasing sequence of 4 numbers(out of numbers 1,2,3,4,5) >1111 1112 1113 1114 1115 >1222 1223 1224 1225 1233 >1234 1235 1244 1245 1255 >1333 1334 1335 1344 1345 >1355 1444 1445 1455 1555 >2222 2223 2224 2225 2233 >2234 2235 2244 2245 2255 >2333 2334 2335 2344 2345 >2355 2444 2445 2455 2555 > >are only 45 options. > >You can translate every non decreasing sequence by adding 1 to the second >number, adding 2 to the 3th number and adding 3 to the 4th number. > >The problem is how many increasing sequences you can choose from the numbers 1-8 > and the answer is 8!/(4!*4!)=70 > > >Uri
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