another simple way to look at it is

lowest number possible 0000
highest number possible 9,999
0000 to 9,999:  thats 10,000 numbers

> > I'm design an electronic for a 16f84 and I'm wondering
> > what's the correct way to determine the total number of
> > possible different combinations that could be input
> > before finally obtaing the correct code.
> >     My keyboard has 10 digits ( 0 - 9 ) and I've chosen
> > to have a 4 digit code to open the lock. I believe the
> > correct way to determine the total combinations possible
> > would be exactly 10000 ( i.e. 10^4 = 10000 ) but, I've
> > been told the proper way to determine this is to use
> > factorials ( i.e. 10! / 4!  = 151200 ).
>
>      So I guess there is a reason they make us EEs take discrete
> probability... ignore the "advice". Your problem is one of the more simple
> problems, I always think of it this way. You have 4 spots, and each spot
you
> can choose 10 different objects. So you have 10x10x10x10 possibilities, or
> 10^4. The 10!/4! thing deals with something slightly different (it deals
> with no repetition allowed, and it's still wrong, should have been
10!/6!),
> from my Discrete Mathematics text:
>
> r-permutation   Repetition not Allowed  n!/(n-r)!
> r-combination   Repetition not Allowed  n!/((n-r)!r!) (the lottery)
> r-permutation   Repetition Allowed      n^r (what you wanted)
> r-combination   Repetition Allowed      (n+r-1)!/((n-1)!r!)
> (From Discrete Mathematics and Its Applications by Kenneth H. Rosen, 4th
> Edition, Page 291)
>
> Hope this helped, TTYL