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Topic: unique ordering
Replies: 9   Last Post: May 10, 2014 1:16 PM

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scattered

Posts: 85
Registered: 6/21/12
Re: unique ordering
Posted: May 10, 2014 9:23 AM
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On Saturday, May 10, 2014 8:41:55 AM UTC-4, scattered wrote:
> On Saturday, May 10, 2014 8:21:28 AM UTC-4, radam...@gmail.com wrote:
>

> > Thought experiment:
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> > Lets say that I randomly draw N letters out of a Scrabble letter-bag. I will then place those letters in a horizontal line on a table. How many UNIQUE patterns of letters can I produce? If there were no duplicates it would be easy (N factorial) but when there are duplicate letters it seems more complicated.
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> >
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> > Bob
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> For any specific set of N letters it is trivial:
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> N!/((#A)!(#B)!...(#Z)!(#Blanks)!)
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>
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> where #A = the number of A's occurring in the string, etc.
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> Most factors in the denominator are of the form 0! = 1 or 1! = 1 and can be
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> omitted.
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> For example there are 11!/(4!4!2!) distinguishable permutations of
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> MISSISSIPPI since #I = 4, #S = 4, #P = 2
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> If you are talking about random collections of N tiles drawn from a
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> standard set of Scrabble tiles then the number of distinguishable
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> permutations becomes a random variable. It would most likely
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> require a computer to enumerate the range of this random variable,
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> calculate its expected value, etc.


I wrote a Python script to perform the experiment of randomly
drawing N scrabble tiles 100,000 times and estimating the expected
number of distinguishable permutations. The output for N in the range
1 to 10 was:

1: 1.0
2: 1.95047
3: 5.56239
4: 20.63273
5: 93.27175
6: 494.08351
7: 2987.684
8: 20141.765
9: 149311.79592
10: 1201688.8506



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