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From: Nigel Wade <nmw@ion.le.ac.uk>
Newsgroups: comp.lang.java.programmer
Subject: Re: Mathematics, Java, good enough
Date: Tue, 17 May 2016 12:56:32 +0100
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On 17/05/16 11:49, Duncan Idaho wrote:
> Hello again
>
> I have a question that involves mathematics. I agree that it's off topic but as I am writing in Java
> I thought I'd give it a go (tenuous I know but it's the best I've got).
>
> I'm dredging up set theory from the furthest reaches of my memory. I'll probably get it wrong.
> Homework? If only, that would mean I could have the last 50 years again ... perhaps not. %-)
>
> Anyway
>
> I have a set of characters S of cardinality C
> example
> if S = {a..z, 0..9, A..Z} then C = er, um  ..,62.
>
> For any given result string X, X need not be a set.
>
> eg if L = 8 then
> "abcdefgh", "AAAAAAAA", "AABBCCDD" are all valid values for X
>
> how many strings of length L are possible from S
>
> I don't even know what to call the question and I probably won't understand the answer but it will be fun trying.
>

The answer is pretty simple. 62^8.
The reasoning is also pretty simple.

The first character has 62 possibilities.
The second character also has 62 possibilities, so the first 2 characters have 62*62 possibilities (62^2)
The third character also has 62 possibilities, so the first 3 characters have 62*62*62 possibilities (62^3)
In general, for L from S, the number of possibilities is S^L.

If you are not convinced, then just remove the letters and work with the numbers.
The first digit can be 0-9. The second digit can be 0-9, so the allowed values are 00-99. And so on. This ought to be 
quite familiar. For 8 digits the range of possibilities would be 00000000-99999999.


I'm not sure of the specific name is, most likely combination of probabilities.