No one asked what I was doing with 120,526,770 records

[Jay R. Ashworth]

On Sun, Feb 13, 2005 at 09:52:56AM -0500, Kenneth Brody wrote:
> Quoting "Jay R. Ashworth" <jra at baylink.com>:
> [...]
> > Unless the powerball works differently than most multiball lotteries, I
> > suspect you're wrong.
> >
> > [ reads powerball.com ]
> >
> > Nope.  It's 53*52*51*50*49 * 42
> >
> > Why are you dividing?
> >
> > The first five balls come from a descending pool of available balls;
> > the last one from a separate pool.
> >
> > So, the correct numeric answer, actually, is 14,463,212,400, rather
> > than what I posted before.  But still a vastly larger number than
> > yours.
> [...]
> 
> You want _combinations_, not _permutations_.  Unless this powerball
> requires that you pick the numbers in the _same_order_, you need to
> divide by 5! for a 5-ball game.
> 
>     14,463,212,400 / 5! = 120,526,770

It's not a permutations thing.  It's a "the ball you just picked isn't
in the hopper anymore" thing.  The first pick is 1 ball in 53.  The
second pick is therefore 1 ball in 52, and so forth.  IE: the Factorial
already *handled* that issue, and you don't need to do it again.

I may still be wrong, but I'm afraid it's going to take a bigger 2x4 to
explain it to me than that, Ken.  :-)

Cheers,
-- jra
-- 
Jay R. Ashworth                                                jra at baylink.com
Designer                          Baylink                             RFC 2100
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      If you can read this... thank a system adminstrator.  Or two.  --me