on facets and unordered combinations

I have to come up with some sort of basic taxonomy for our files at work, so today I was doing some research to help my mind get where it needs to be. First, I need to mention this [http://www.taxonomystrategies.com/presentations/Getting_Started.ppt] really nice PowerPoint on developing a taxonomy for work called "Getting Started with Business Taxonomy Design". It got me thinking about tags and how if I tagged something with four tags, "a", "b", "c", and the equally dull "d", then I could calculate the number of ways I could filter down to a set of results that would contain that document. I don't know if my thinking is correct, but I'm thinking there should be 15 ways to get to that document with four tags because I should be able to filter, inclusively, not only by each individual tag but also by unordered combinations of those tags, i.e. tagged with both "b" AND "a", etc. So, I read this [http://www.mathsisfun.com/combinatorics/combinations-permutations.html] neat page on calculating unordered combinations for a set and wrote a Python script, below, to calculate the total. I had to also see this [http://www.adonald.btinternet.co.uk/Factor/Zero.html] page about what the value of zero factorial is as I'd assumed it would be undefinable. The script tells how many unordered combinations for four tags there would be for a group of four tags (i.e. all of them), then three, then two, then each of the four. import math tags = ['a','b','c','d'] #say I have assigned 4 tags to a file tags_len = len(tags) numerator = math.factorial(tags_len) combinations_value = 0 i = tags_len def calculateCombinations(i): denominator = (math.factorial(i)) * (math.factorial(tags_len-i)) value = (numerator/denominator) return value while i > 0: print calculateCombinations(i), 'unordered combination(s) per', i combinations_value = combinations_value + calculateCombinations(i) i = i - 1 print '-----\n=', combinations_value, 'unordered combinations' Here's the output: >>> <br/> 1 unordered combination(s) per 4<br/> 4 unordered combination(s) per 3<br/> 6 unordered combination(s) per 2<br/> 4 unordered combination(s) per 1<br/> -----<br/> = 15 unordered combinations<br/> >>> <br/> And here's me manually calculating it just to be sure ... QUADS (1):<br/> abcd<br/> <br/> TRIPLES (4):<br/> abc; abd; acd<br/> bcd;<br/> <br/> DOUBLES (6):<br/> ab; ac; ad<br/> bc; bd<br/> cd<br/> <br/> SINGLES (4):<br/> a<br/> b<br/> c<br/> d<br/> -----<br/> 1 + 4 + 6 + 4 = 15 unordered combinations<br/>