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- ...he six men. so the final answer would be P(10,10)x P(11,6) --[[User:Krwade|Krwade]] 20:52, 4 February 2009 (UTC)2 KB (390 words) - 21:41, 4 February 2009
- my email is krwade@purdue.edu187 B (30 words) - 18:06, 27 January 2009
- ...en subtract that from the total ways with a which is 26^6. --[[User:Krwade|Krwade]] 21:11, 4 February 2009 (UTC)1 KB (193 words) - 21:45, 4 February 2009
- ...and you multiply the three together...e and f are similar --[[User:Krwade|Krwade]] 21:33, 10 February 2009 (UTC)1 KB (198 words) - 19:40, 11 February 2009
- ...e the C(80,11)? It just seems like that number is important--[[User:Krwade|Krwade]] 16:41, 17 February 2009 (UTC) I guess I still have the same question as Krwade. Where does the number C(80,11) come into play? It does seem like it will3 KB (433 words) - 18:51, 19 February 2009
- ...be: <math>\frac{{6 \choose 5}(34)}{{40 \choose 6}}</math> --[[User:Krwade|Krwade]] 16:55, 17 February 2009 (UTC) ...ill be <math>\frac{{6 \choose 5}{34 \choose 1}}{{40 \choose 6}}</math>-- Krwade is right..2 KB (252 words) - 15:43, 18 February 2009
- Does anyone understand part C. It's confusing me. --[[User:Krwade|Krwade]] 20:48, 24 February 2009 (UTC)629 B (91 words) - 15:23, 25 February 2009