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Enneagram Type 5 Board Archive Re: But...ummmPosted by Bartholomew on January 23, 2002 at 18:58:49: In Reply to: Re: But...ummm posted by Mikko on January 23, 2002 at 07:11:31: : --I did some tinkering of my own, I don't know if it makes any sense but at least the end result is interesting. Here's the way I see it: : The way Honcho solved the problem would require there are 4 different categories of digits, instead of two. It would require that only in his first case. In his second through fourth cases, he has certain digits the same, and if there were 4 different categories of digits this would not be possible. : In that case using 4! (or 4!/1!) for the number of possible combinations would be correct. Since there are only two categories of digits, each occupying two positions, you have to use 4!/2! = 6 4!/2! = 24/2 = 12. : in order to avoid counting those certain possibilities twice. (ie the permutation of the total number of positions divided by the permutation of the number of categories) If you apply that figure to the method Honcho used, you'll get : (52*51*10*9*6) + (52*51*10*6) + (52*10*9*6) + (52*10*6) = 1622400 : If you had 3 numbers and one letter, it would be 4!/3! and all numbers it would be 4!/4! : I think it doesn't, because you screwed the arithmetic up and still got the right answer. If you had not screwed the arithmetic up, you would have gotten the wrong answer, so your concept is flawed. I couldn't give you a more in-depth analysis than that without reviewing basic probability, which I only learned to do well on the SAT II: Mathematics IIC, which I took LAST year. I've forgotten the formulas by now through disuse. The way I considered it was that for each pair of numbers/letters, half of them would be the same as the other half, only reversed - like (5, 9) and (9, 5), or (a, e) and (e, a). So really only half of the number pairs and half of the letter pairs should be counted. So I would say (for Honcho's first case), ((52*51)/2)*((10*9)/2)*24. That seems simpler.
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