August 24th, 2015, 04:08 PM  #21  
Newbie Joined: Aug 2015 From: Beijing, China Posts: 3 Thanks: 4 Math Focus: Number theory? Do I even know what that is?  Quote:
ex: if you have many options (>=44) you can go 12345, 12678, 1291011, ... , 12 3k  3k+1  3k+2 , ... but for a group of tuples that not everyone share the same pair, i.e. not the case above, think about why a pair cannot appear more than 4 times in all tuples. What kind of bad thing will happen? Similarly, think about why a pair cannot appear exactly 3 times while also have tuples not containing that pair. If you see why a pair cannot appear for more than 4 times, then imagine choosing the first tuple, the choice of which is immaterial. Suppose it's 12345. Can you have more than 31 tuples in total, if each of the new tuple have to share a pair with 12345, yet each pair in 12345 cannot appear for more than 4 times? Having 14 options and having 14000000 options yield the same answer. Using this, you can shrink the case with each pair appearing at most twice to 11 and shrink the case where each pair can appear four times to 13. The case when everyone share the same pair is relatively simple.  
August 25th, 2015, 08:49 AM  #22  
Banned Camp Joined: Dec 2013 Posts: 1,117 Thanks: 41  Quote:
Please can you explain it giving examples?  
August 25th, 2015, 07:41 PM  #23  
Newbie Joined: Aug 2015 From: Beijing, China Posts: 3 Thanks: 4 Math Focus: Number theory? Do I even know what that is?  Quote:
Like if you changed the number of possible symbols from 11 to 12, the maximum number of tuples you can form is still 11.  
August 27th, 2015, 08:30 AM  #24 
Banned Camp Joined: Dec 2013 Posts: 1,117 Thanks: 41 
I abandon this thread. I came to the conclusion that combinatorics as long as it is hard few will be interested in. Thank you all for trying! 
August 27th, 2015, 10:13 AM  #25 
Senior Member Joined: Dec 2007 Posts: 687 Thanks: 47  You got answers, it is not that hard, it is perhaps not so interesting as CRG said, especially if it is not related to any other nice result or open problem.

August 27th, 2015, 10:23 AM  #26 
Banned Camp Joined: Dec 2013 Posts: 1,117 Thanks: 41 
For sure I got answers but not what I really needed. I thanked you for this. I did not publish more algorithms since no one tried those published. I`m not programmer. Starting from today I will not post any subject related to combinatorics. All the results I have from posters are not new for me. It is not only here but in other forums too. Number theory and combinatorics finished! no more posts about those 2 subjects. Anyway thanks lot. 
August 27th, 2015, 10:38 AM  #27 
Senior Member Joined: Dec 2007 Posts: 687 Thanks: 47  
August 27th, 2015, 10:54 AM  #28  
Global Moderator Joined: Nov 2006 From: UTC 5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms  Quote:
If someone does, though, it would probably merit a new sequence at the OEIS.  
August 27th, 2015, 11:23 AM  #29 
Banned Camp Joined: Dec 2013 Posts: 1,117 Thanks: 41 
I have designed until now 6 algorithms to solve this problem. I have published on internet 3 of them. I did not abandon the problem. I`m still thinking to solve it. But because I`m not programmer I will solve it with pen and paper. I have hard time to focus because of health problems. Anyway thank you all. 

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combinatorics, hard 
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