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: > > > > > > > > > > > > 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.
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: