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 February 14th, 2015, 11:30 AM #1 Newbie   Joined: Feb 2015 From: washinton state Posts: 11 Thanks: 0 Math Focus: limited knowledge. but interested in computer science and permutations. "for my redstone creations" permutations if i want to know the actual combinations how would i do that. i have "1,2,3" the formula gives me 15 combinations based on three subsets. so i now know the number of combinations. however i dont know the actual combinations. for a three number system its easy. subset one (1.2.3) (3,2,1) (3,1,2) (2,1,3) (2,3,1) (1,3,2) subset two (1,2) (2,1) (2,3) (3,2) (3,1) (1,3) subset three (1) (2) (3). the actual combinations is what im after. is there an easy way to be sure that i got ever combination. because unlike with three (which has 15 combinations) if i do it with "1,2,3,4" i have 64 combinations spreed across 4 subsets. i understand in this instance ("1,2,3,4)" each number will occur 6 times per "vertical space" for the first subset. but i get lost and if i do this with 1-10 well thats a lot to keep track of. i am trying to find a way to organize the numbers to get all combinations so all i have to do is check two directions to confirm my permutations. 1 2 3 4 4 1 2 3 3 4 1 2 2 3 4 1 this makes a line vertically and diagonally. i get this far then get lost where to pick up form there. each number "if im understanding this correctly" must occur six times in each of the four vertical rows. i will accept a website that has a calculator to find them, however i would prefer to make a chart. i like finding and mastering patterns. additional thought if i continue the above chart would starting with a new "first" number solve my problem or just part of it? 2 1 3 4 4 2 1 3 3 4 2 1 1 3 4 2 and then repeat a new "first" number with each available number. does that ensure every combination resulting from the permutation is counted?
 February 14th, 2015, 12:31 PM #2 Global Moderator     Joined: Nov 2006 From: UTC -5 Posts: 16,046 Thanks: 933 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms This is essentially just writing in binary. If 1 means an element is in the set and 0 means it isn't, the 8 (7 nonempty) subsets are 000, 001, 010, 011, 100, 101, 110, and 111.
 February 14th, 2015, 12:45 PM #3 Newbie   Joined: Feb 2015 From: washinton state Posts: 11 Thanks: 0 Math Focus: limited knowledge. but interested in computer science and permutations. "for my redstone creations" i do know that however, how do i know if i got all 64 combinations associated with "1,2,3,4" that is my problem i could do 0001,0010, 0100, 1000, however i dont need to repersent zero therefor i would end up with extra combinations. to cary on 0004, 0040, 0400, 4000, then more so 1234, 4123, 3412, 2341, which just bring me back to my stated problem. how do i know i got them all or basically when does the pattern end?
 February 14th, 2015, 12:51 PM #4 Math Team   Joined: Dec 2006 From: Lexington, MA Posts: 3,267 Thanks: 407 Hello, martinthewarrior! With $n$ digits, I suggest making 1-digit numbers, 2-digit numbers, 3-digit numbers, . . . $n$-digit numbers, and listing them in increasing order. $\text{For }n=3:\;\{1,\,2,\,3\}$ $\;\;\;\;\;\text{ we have: }\;\;\begin{array}{c}1\\2\\3\end{array}\;\;\; \begin {array}{c}12\\13\\21\\23\\31\\32\end{array} \;\;\;\begin{array}{c}123\\132\\213\\231\\312\\321 \end{array}$
 February 14th, 2015, 01:57 PM #5 Newbie   Joined: Feb 2015 From: washinton state Posts: 11 Thanks: 0 Math Focus: limited knowledge. but interested in computer science and permutations. "for my redstone creations" i have that example in my statement but i still dont know how to be sure i got all the combinations. i need it for 1,2,3,4,5 which means there is 265 combinations as with (X,X,X,X,X) (X,X,X,X) (X,X,X) (X,X) (X) trying to find all combinations and figuring out which are still need within a subset, for example: 1,2,3,4,5 after i get to the 25th pairing i get confused or lost as to what combination i need next or have already done. there are 120 ways to configure (1,2,3,4,5) another 120 to do (X,X,X,X) 60 to do (X,X,X) 20 to do (X,X) and 5 to do (X) thats a lot to keep track of i already split up each subset as you suggested but 120 combinations is a lot and i need a way to ensure i got each variant. and did not copy one agin or get to the end and find some missing.
February 14th, 2015, 02:08 PM   #6
Math Team

Joined: Apr 2010

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Quote:
 Originally Posted by martinthewarrior i need it for 1,2,3,4,5 which means there is 265 combinations as with (X,X,X,X,X) (X,X,X,X) (X,X,X) (X,X) (X)
As you find below, it would be 120 + 120 + 60 + 20 + 5 = 325 combinations.

 February 14th, 2015, 02:23 PM #7 Newbie   Joined: Feb 2015 From: washinton state Posts: 11 Thanks: 0 Math Focus: limited knowledge. but interested in computer science and permutations. "for my redstone creations" oh oops i missed typed i guess. but to show you my problem ill show you what ive been doing. this is for the third subset the first two are easy but i get lost here. im doing this for 1 2 3 4 5 1 2 3 1 3 2 1 4 2 1 2 4 1 3 4 1 5 2 1 2 5 1 3 5 1 5 3 ok the bold "2s" are a pattern that reoccurs for the numbers 2, 3, 5 but not 4. 1 is excluded because its constant. it reoccours for 3 because it repeats back to the top which made me think it was a complete cycle, however i now see 4 is ot fallowing the pattern. so i figured since im combining three number together (odd) that its not going to fallow and even pattern which the pattern is. so unless im misunderstanding something i just fixed my problem. unless you still have difficulty understanding what i am saying. Last edited by martinthewarrior; February 14th, 2015 at 02:54 PM. Reason: re explained
 February 14th, 2015, 02:41 PM #8 Math Team   Joined: Apr 2010 Posts: 2,778 Thanks: 361 Sorry, I don't really know what you mean. For starting, I don't know what 'that is repeated for 2 4 5 but not 3.' means. I can think of an approach for your problem, but I'm not sure if it would help you. I doubt, because I could give a PARI-prog, but maybe you don't want to use it and maybe you want to do it pen and paper. The approach is as follows: An example, to get the four-digit numbers, first, compute all the permutations of the integers [1,2,3,4]. There are 24. Then, find all vectors having binary elements (0 or 1) and the elements in ascending (or descending order). There are 5 (for both ascending and decending). Sum any of the permutations of [1,2,3,4] with any of the 'ascending' vectors and you get one of the numbers you want. There are a total of 24 * 5 = 120 such vectors, precisely those you want. What do you think?
 February 14th, 2015, 02:44 PM #9 Newbie   Joined: Feb 2015 From: washinton state Posts: 11 Thanks: 0 Math Focus: limited knowledge. but interested in computer science and permutations. "for my redstone creations" ill restate the confused part about the repeat for 245 not 3 part.
 February 14th, 2015, 03:01 PM #10 Math Team   Joined: Apr 2010 Posts: 2,778 Thanks: 361 I think, I see what you mean. 2 and 5 show a pattern somewhere in the list of numbers where the pattern is that in 4 consecutive numbers, the occur second, than third, than third again, than second. This is seen in the first four numbers for 2 and the last two you listed for 5. Could this help you find all the numbers you search? I don't know, but you seem to have solved it. Haven't you?

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