Re: A83: Tutorials...

To: assembly-83@lists.ticalc.org
Subject: Re: A83: Tutorials...
From: Bryan Rabeler <brabeler@isd.ingham.k12.mi.us>
Date: Tue, 02 Jun 1998 15:36:17 -0400
Delivered-To: assembly-83-outgoing@towerguard.unix.edu.sollentuna.se
References: <yam7457.1268.275600792@smtp.softhome.net>
Reply-To: assembly-83@lists.ticalc.org


Linus Akesson wrote:

> On 01-Jun-98, Bryan Rabeler wrote:
>
> >James Matthews wrote:
>
> >> PS:  Bryan, Harper (the guys with the connections...:):  is there anyway I
> >> can get my tutorials REVIEWED on either ticalc or Ti-Philes...I want 1. the
> >> publicity :), and 2. Constructive feedback...Conway was the only guy who
> >> gave me feedback who I didn't talk to personally (over ICQ...like Harper
> >> and Joe).  Thanks.
>
> >You can submit them to features@ticalc.org to be published in the articles
> >section.
>
> >> PPS:  I failed my math final :)...there was a question on "mutual
> >> exclusive"...if you have the number 341521 how many ways can you make a odd
> >> number if you re-arrange numbers.  360...please say 360 :)...
>
> >There are 3 different odd numbers in that 6 digit number, so there are 3
> >possibilites for the last place value.  The other place values don't matter.
> >So there are 5 posibilites for the first place value, 4 for the second, 3 for
> >the third, 2 for the fouth, 1 for the fith, and 3 for the last one.
> >5*4*3*2*1*3 = 360.  But since there are two 1's, you must divide that by 2
> >because you could swap the 1's and you would get the same number.  So I think
> >the answer should be 180.
>
> >--
> >Bryan Rabeler <brabeler@ticalc.org>
> >   File Archives, HTML, and Support
> >   the ticalc.org project - http://www.ticalc.org/
>
> Hmm.. 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

If the last digit is 1, then the number of odd numbers is:

5*4*3*2*1 = 120

If the last digit is 3 or 5, then the number of odd numbrs is:

5*4*3*2*1/2 = 60 (divide by 2 to cancel out the duplicates caused by two 1s)

120 + 60 = 180.

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

References:

Re: A83: Tutorials...

From: Linus Akesson <lairfight@softhome.net>