Re: A83: Tutorials...

[Prev][Next][Index][Thread]

Re: A83: Tutorials...


Icemn711@aol.com wrote:

> That is the right way.
>
> >  don't you pay too much attention to those 1s now? First you say that
> >  only 3 digits can be the rightmost digit. Then when you've calculated the
> >  number of possible combinations, you divide that number by two. But by then
> >  you've already reduced the number of possible combinations once! I'd say 4
> >  digits can be in the units place (3, 1, 5 and 1), then calculate
> 5*4*3*2*1*
> > 4
> >  = 480. _This_ number should be halved -> 240 possible combinations.
> >
> >  Am I completely wrong?
> >
> >  Linus

112345    112435    112453    112543    113245    113425    114235    114253
114325    114523    115243    115423    121345    121435    121453    121543
123145    123415    123451    123541    124135    124153    124315    124351
124513    124531    125143    125341    125413    125431    131245    131425
132145    132415    132451    132541    134125    134215    134251    134521
135241    135421    141235    141253    141325    141523    142135    142153
142315    142351    142513    142531    151243    151423    152143    152341
152413    152431    153241    153421    154123    154213    154231    154321

Ok, the above are all the posibilites for having a 1 in the first place value.
(tell me if I missed any)  There are 64 total odd numbers.  For each number, you
could swap the 1 in front with any of 4 other different numbers, so there should
be 64*5 = 320 different numbers.

--
Bryan Rabeler <brabeler@ticalc.org>
   File Archives, HTML, and Support
   the ticalc.org project - http://www.ticalc.org/
References: