Re: A83: Tutorials...
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Re: A83: Tutorials...
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
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