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/