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Let's say I have 5 variables (a,b,x,y,z).

  • Each variable $\in (1,2,3,4,5,6,7,8,9,10,11,12,13,14,15)$.
  • The length of each variable must be 3. So a can be e.g. ${1,2,3}$ or ${2,3,4}$ or ${3,4,5}$, etc...
  • The numbers is each variable must be consecutive e.g. (5,6,7) or (11,12,13)

My question is how do I calculate the possible variations of all 5 variables combind resulting in the above. Basically, what's the number of possible combinations of n number of variables that satisfy the above rules.

This is my very first question on the math site, so im definitely not using the right terminology or tags to describe the question

Yehia A.Salam
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  • I mostly am thinking, what has been tried ? Also, do you mean consecutive, when you say sequence numbering? Because technically 1,3,5 is an arithmetic progression, but not consecutive. Can variables overlap ? all these questions answers, may change the answer. I think you could add more tags but I don't agree that yours is a wrong tag. –  Jul 18 '17 at 01:01
  • possibly related and or duplicate of : https://math.stackexchange.com/questions/1611927/how-to-compute-for-the-number-of-sequential-combinations-possible-in-a-set?rq=1 –  Jul 18 '17 at 01:04
  • @RoddyMacPhee i updated the question, they are consecutive with no overlap – Yehia A.Salam Jul 18 '17 at 01:04
  • @Roddy it's the same actually, except here i'm trying to find the number of combinations of the different variation for each variable , i guess multiplying them would do, right ? – Yehia A.Salam Jul 18 '17 at 01:08
  • do they wrap around like (14,15,1) ? –  Jul 18 '17 at 01:10
  • nop no wrapping – Yehia A.Salam Jul 18 '17 at 01:10
  • Let us continue this discussion in chat. –  Jul 18 '17 at 01:15

1 Answers1

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Given what was told in the comments and question, it follow that:

  1. there are 13 possible consecutive sets for each variable$$\{\{1,2,3\},\{2,3,4\},\{3,4,5\},\{4,5,6\},\{5,6,7\},\{6,7,8\},\{7,8,9\},\{8,9,10\},\{9,10,11\}\{10,11,12\}\{11,12,13\}\{12,13,14\}\{13,14,15\}\}$$
  2. with no overlap the first variable can be any one of these 13, the second can be 12, of them the third 11,fourth 10,fifth 9 and you are right we multiply these to get: 154440 . Then, we realize that since ordering of the variables, isn't considered important, then we can divide by 5!=120 to get 1287.
  3. Of course, this doesn't consider, the no overlap in values. Once we do that, we have that there is only one way ( without ordering becoming important) that this can be done. $$\{\{1,2,3\},\{4,5,6\},\{7,8,9\}\{10,11,12\},\{13,14,15\}\}$$
  • actually the second and third and fourth (etc..) variable can have the same sets. So variable y can have {1,2,3}, while x have {1, 2, 3} as well. So according to the linked answer as well, the total number of possibilities would be $K−(N -1)$ where K is the number of elements in the set (15) and N is the length (3). And then muliplying $K - (N - 1)$ * $K - (N - 1)$ * $K - (N - 1)$ I would get the total possibility for three variables = 13* 13 * 13 = 2,197, right? – Yehia A.Salam Jul 18 '17 at 22:40
  • I'm confused why you said no overlap then I guess. I might be back later. –  Jul 18 '17 at 22:59
  • sorry i meant they don't wrap around, like 14,15,1 – Yehia A.Salam Jul 18 '17 at 23:00
  • what about ${1,2,3};and;{2,3,4}$ ? –  Jul 18 '17 at 23:03
  • yea that one is possible – Yehia A.Salam Jul 18 '17 at 23:15
  • then $$13^n$$ for n variables would work –  Jul 18 '17 at 23:19