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#1 2006-12-29 22:42:54

Toast
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Registered: 2006-10-08
Posts: 1,321

Disregarding repeated objects

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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#2 2006-12-30 03:30:33

krassi_holmz
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Registered: 2005-12-02
Posts: 1,905

Re: Disregarding repeated objects

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

elements, in which the i-th element has
copies, is:


IPBLE:  Increasing Performance By Lowering Expectations.

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#3 2006-12-30 03:58:26

Toast
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Registered: 2006-10-08
Posts: 1,321

Re: Disregarding repeated objects

Yeah, but I'm not dividing by the repeated objects until I actually get my answer from non-repeated objects

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#4 2006-12-30 04:06:15

krassi_holmz
Real Member
Registered: 2005-12-02
Posts: 1,905

Re: Disregarding repeated objects

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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