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Say if you have 10 objects, A A A B B B C D D E, and you had restrictions (say you must have a vowel at the ends or the vowels must be together), could you just work through the problem as though each letter was different, then divide the final answer by the repeated letters?
You for instance pretend the vowels are instead A E I O, and the consonants are B D F G H J, then by placing the vowels together get 4!*7!=120960 combinations, then divide the answer by 3!3!2! (repeated letters) to get 1680.
Does this work for every question? Because then I'll probably use this for every question.
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Yes. That's because every normal word with n A's is equivalent to n! "non-normal" words (in which we use A1,A2,A3,...,AN instead of A,A,A...,A), because of replacing the A's, if you understand me.
There is a general result:
The number of all permutations of
IPBLE: Increasing Performance By Lowering Expectations.
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Yeah, but I'm not dividing by the repeated objects until I actually get my answer from non-repeated objects
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Yes. You first get the answer for the non-repeating objects, then divite to obtain the repeated objects. That's a good technique, because often the number of non-repeating objects with certain properties is easier to calculatre than calculating directly for repeating objects.
Last edited by krassi_holmz (2006-12-30 04:06:48)
IPBLE: Increasing Performance By Lowering Expectations.
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