Dealing with Duplicate Elements
Date: 05/17/2002 at 09:08:52 From: Laura Subject: counting techniques - probability I have a question that requires using factorial notation to calculate how may different words can be formed in each of the following cases; a) BOOK - I know there are 12 words, but how do I write this with factorial notation? b) ASSIST - ? c) SERIES - ? d) MATHEMATICS - ? Then I need to develop a general formula to calculate the number of different "words" that can be formed from an N letter word with n1, n2, n3, ....nk letters which are the same. I experimented with a different word lengths, containing 2 of the same letters, as I thought this may help, e.g., EEL - 3 words WOOL - 12 words STOOD - 66 words I tried to find a pattern, but couldn't, so I'm stuck Any help would be appreciated! Thanks.
Date: 05/17/2002 at 09:37:41 From: Doctor Mitteldorf Subject: Re: counting techniques - probability Dear Laura, You're doing the right thing, but it may be hard to guess the pattern. I won't give away the fun by telling you the answer, but I'll give you a hint as to how to think about it: First, count the number of permuatations assuming all the letters are different. I think you may already have that answer. Then consider how that answer changes if there is a single pair of letters that is the same. I think you may already have that answer in mind as well, but it will help to articulate it and write it down formally. Next, think what happens if there are two such pairs, as in the word LETTER. When you've got that clearly in mind, the next step is to do a single triple letter, as in the word POPPY. This may be a signficant intellectual leap, but once you have it, you'll have the key for generalizing the result. Hint: The principle is to think about the subset of letters that are identical, how many ways they can be invisibly rearranged among themselves. - Doctor Mitteldorf, The Math Forum http://mathforum.org/dr.math/
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