Re: A83: Tutorials...

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Re: A83: Tutorials...


Since you have a six digit number with three (not including the second one)
odd numbers, wouldn't you just do 6!*(3/6) and get 360?  Or am I completely
off?
Paul

brabeler@isd.ingham.k12.mi.us wrote:

> 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>
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