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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 8:41 AM
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On Saturday, May 10, 2014 8:21:28 AM UTC-4, radam...@gmail.com wrote:
> Thought experiment:
>
>
>
> 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.
>
>
>
> Bob


For any specific set of N letters it is trivial:

N!/((#A)!(#B)!...(#Z)!(#Blanks)!)

where #A = the number of A's occurring in the string, etc.
Most factors in the denominator are of the form 0! = 1 or 1! = 1 and can be
omitted.

For example there are 11!/(4!4!2!) distinguishable permutations of
MISSISSIPPI since #I = 4, #S = 4, #P = 2

If you are talking about random collections of N tiles drawn from a
standard set of Scrabble tiles then the number of distinguishable
permutations becomes a random variable. It would most likely
require a computer to enumerate the range of this random variable,
calculate its expected value, etc.



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