If you take the characters a-z and the numbers 1-9, you have 35 characters, yes. When you use them in a sequence of up to 3 characters, you have:
- 35 + 35^2 + 35^3 = 44,135 sequences Then, if you take three of those sequences, and allow repeats, you have - 44,135^3 = 85970488160375 possible triplets If you don't allow repeats, you have: - 44,135 * 44,134 * 44,133 = 85964644553970 possible triplets. At least, that's what I get, using my admittedly rusty knowledge of combinatorics. Thanks, Rohit Patnaik On Sun, Oct 14, 2012 at 5:19 PM, Kevin LaTona <[email protected]> wrote: > I am working on a weekend project idea and was wondering if anyone on the > list could verify a math problem for me? > > > > Here is the problem. > > If one takes the characters a-z and numbers 1-9 this gives one 35 > possible character options. > > Now if used in a sequence of up to 3 characters. > > I get a total of 6,545 possible combinations of these 35 characters. > > > > Now if one was to use them in file folder tree style structure such as, > > ad6 / 7gh / d8s > > My math shows there is 46,706,638,440 possible combinations of these 35 > characters in a 3 layer deep tree. > > Do you get the same results or am I doing something wrong here? > > Thanks your thoughts on the matter. > -Kevin > > > > > This is the rather quick and dirty Python code I used to get to these > results today. > > import math > > a = math.factorial(35) > b = math.factorial(35-3) > c = math.factorial(3) > d = a / (b*c) > '{:,}'.format(d) > #'6,545' > > > e = math.factorial(6545) > f = math.factorial(6545-3) > g = math.factorial(3) > h = e / (f*g) > '{:,}'.format(h) > #'46,706,638,440' >
