Find Number of Permutations of 2D Numbers Constrained by Sum n

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Discussion Overview

The discussion revolves around finding the number of permutations of 2D numbers (Xk) constrained by the condition that their sum equals a natural number n. The context includes mathematical reasoning related to combinatorial approaches and applications in classical dynamics.

Discussion Character

  • Exploratory, Technical explanation, Mathematical reasoning

Main Points Raised

  • One participant introduces the problem of finding permutations of 2D numbers constrained by their sum being n, specifying that the numbers are whole numbers.
  • Another participant notes that permutations of the numbers do not impose additional constraints since any permutation will maintain the same sum.
  • A different viewpoint suggests that if the numbers are non-negative integers, the number of solutions can be determined using combinations with repetition, framing it as filling variables until reaching the total n.
  • One participant presents a visual analogy involving balls and sticks to represent the problem, proposing that the arrangement of these objects can be used to calculate the number of permutations, leading to a combinatorial formula.
  • A later reply expresses gratitude for the explanation, indicating that it addressed a significant concern regarding the problem.

Areas of Agreement / Disagreement

Participants present various approaches and interpretations of the problem, with no consensus reached on a single method or solution. Multiple competing views remain regarding the best way to calculate the permutations.

Contextual Notes

The discussion includes assumptions about the nature of the numbers (whole and non-negative integers) and the combinatorial methods used, which may depend on specific interpretations of the problem constraints.

Ketsha
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ok this seems like a predominantly maths problem but i need it for a little project of mine related to classical dynamics.

So i have 2d numbers Xk, which are constrained by the relation that their sum is n.

what i need to find out is the number of possible permutations of the 2d numbers (Xk) that satisfy this condition.

the numbers Xk belong to the set of whole numbers, and n belongs to the set of natural numbers.
 
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If you have a set of numbers for which x1 + x2 + ... + x2d = n, then any permutation of the xk's will have the same sum, so that isn't much of a constraint.

If you had 4 numbers x1, x2, x3, and x4, whose sum was N, how many permutation of the four numbers are there? Any of the four numbers could go into the first position, then any of the remaining three numbers could go into the second position, then any of the remaining two numbers could go into the third position, leaving only one number to go into the fourth position.
 


Assuming that the numbers are non-negative integers (without this restriction, the number of possibilities is infinite), the number of solutions is given by combinatations with repetition of n objects of k = 2d different types; you may think of this as "filling" each variable Xi, one unit at a time, until you a total of n.

You may find a more complete explanation for these combinations here:

http://www.csee.umbc.edu/~stephens/203/PDF/6-5.pdf"

(These slides are better than the Wiki article)
 
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nice problem...

Consider a number line made of 2d line segments of variable length. each line segment representing one of your 2d numbers. These line segments will be divided by 2d-1 points.. or let's represent by sticks instead of points. so there are 2d-1 sticks.

Now consider N balls. Each representing the numeric value 1. So total number balls is the sum of 2d numbers that you need to be.

now... if you arrange these N balls and 2d-1 sticks randomly, it will be kind off a pictorial representation of your problem. the number of balls between two sticks being a particular value xk.. and there will be 2d such numbers.

Since there can be two suck sticks together... the numbers will be greater than or equal to zero. Hence belonging to the set of whole numbers.

Now the question remains... in how many different ways can we arrange these 2d-1 sticks and N balls.

that will (N+2d-1)!

Now N of them are alike (balls) and 2d-1 of them are alike (sticks). so...

To total number of ways would be... (N+2d-1)!/N!(2d-1)! or simply (N+2d-1)C(2d-1)
 


Thanks for the awesome solution... that was really well explained :) and has solved one of my biggest troubles. :)
 

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