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I need to find the possible permutations (hope that is the right terminology) given a set of items N where order is important and I'm looking for permutations with M number of consecutive items from N while also satisfying the condition that the number of pairings must be P.

An example of this would be having N=4 items 1,2,3,4 and needing to pair these into P=3 groups whil resepcting the order, this would give [1 , 2-3, 4], [1-2 ,3 ,4], [1 , 2 , 3-4].

M here can only be 1 or 2 if I have this right.

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  • $\begingroup$ I don't think you mean permutations. Permutations are are rearrangements, but you always keep the numbers in the same order. I can't understand what you mean by $M$ and $P$. You first say that $P$ is the number of pairings, and then that it is the number of groups. Which is it? What does "$M$ number of consecutive items" mean? In the problem description, it sounds like $M$ is given, but in your example, you says it can be $1$ or $2$. Please try to restate your question more clearly. $\endgroup$ – saulspatz Aug 6 at 15:22
  • $\begingroup$ Thank you for the clarification and the answer, I struggled with putting the probelem into writing. What I meant by P is the number of groups, in my example it is 3 so there has to be 3 groups of items, M I guess is not useful here but what I meant by it is that if we had 3 items in my example and P was 3 then M has to be 1. if N was 5 then M can be 3 since you can have [123,4,5]. M is useful for the specfic application I'm using this for since I want to be able to tune it as well. $\endgroup$ – Alla Aug 6 at 15:26
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If I understand the question what you have to do is choose $P-1$ places in the list $$ 1-2-3 - \cdots - N $$ where you change the $-$ to a $,$ to build the list of groups (in your notation).

So the answer is the binomial coefficient $\binom{N-1}{P-1}$

For example, if $P=4$ and $N=7$ there will be $\binom{6}{3} = 20$ ways. One of them is $$ 1 - 2 , 3 - 4 - 5 , 6, 7 $$

You can look up binomial coefficient in wikipedia.

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  • $\begingroup$ Thank you Ethan, I tried to look into this before posing here but I was a bit overwhelmed, I will read up on binomial coefficients. this does seem to answer the question, I'll just be left with the task of listing those possible ways. $\endgroup$ – Alla Aug 6 at 15:38
  • $\begingroup$ @Alla Many computer languages have packages with methods that cab help with this. $\endgroup$ – Ethan Bolker Aug 6 at 15:47
  • $\begingroup$ Thank you for your valuable input, I will look into those as well and I'm assuming adding the constraint of M is not possible or at least difficult, as in specifying that a group can not exceed M number of items, in your example setting M at 2 making the 2-3-4 group invalid. $\endgroup$ – Alla Aug 6 at 15:50
  • $\begingroup$ @Alla With a maximum of $M$ in each group I think the problem is more difficult. (Your question did not make the meaning of $M$ clear.) $\endgroup$ – Ethan Bolker Aug 6 at 15:54
  • $\begingroup$ That is correct, I did not phrase it well, nonetheless your input has been valuable and I have an idea where to look for now. $\endgroup$ – Alla Aug 6 at 15:55

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