Interesting seems you just gave me plenty of extra headroom with this
idea.
Thanks all for your thoughts.
-Kevin
On Oct 14, 2012, at 5:30 PM, Rohit Patnaik 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'
