Ah yes...I missed the "up to" -- I just looked at the examples and saw that there were always 3 characters.
-- Nate On Sun, Oct 14, 2012 at 5:30 PM, Rohit Patnaik <[email protected]> wrote: > 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' >> > >
