David,
With the original idea I am exploring order does matter.
And repeats on any given tree level is not allowed.
Thanks for your thoughts.
-Kevin
On Oct 16, 2012, at 11:02 AM, David Goldsmith wrote:
Rohit provides one of two crucial distinctions that you did not
provide in your original problem statement: repeats allowed (=>
"exponential counting") or no ("factorial counting")? The other
unprovided crucial distinction: order matters (=> "permutations") or
no ("combinations"), i.e., is "abc" to be considered the same as
"bca" or not?
DG
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'
>
