Number of ways in a 3D lattice

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SUMMARY

The discussion focuses on calculating the number of paths in a 3D lattice, extending from the established 2D case. The correct formula for the number of paths between two points A and B in a 2D lattice is C(n+m, m), where n and m represent the steps in different directions. For a 3D lattice, the number of paths can be generalized using combinations, specifically through trinomial functions, to account for additional dimensions. Participants also referenced external resources, such as Khan Academy videos, to further explore these concepts.

PREREQUISITES
  • Understanding of combinatorial mathematics, specifically combinations (C(n, k))
  • Familiarity with lattice structures in geometry
  • Basic knowledge of trinomial functions and their applications
  • Experience with dimensional analysis in mathematical contexts
NEXT STEPS
  • Research the application of trinomial functions in combinatorial path counting
  • Explore advanced combinatorial techniques for higher-dimensional lattices
  • Study the implications of C(n+m, m) in various mathematical problems
  • Watch the Khan Academy video on 3D path counting for visual understanding
USEFUL FOR

Mathematicians, educators, students studying combinatorics, and anyone interested in advanced pathfinding algorithms in multi-dimensional spaces.

nickek
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Hi!
If I have points A and B in a lattice in the plane, and the closest path between them is n + m steps (for example 4 steps upwards and 5 steps to the right), there are C(9,(5-4)) = 9 combinations of paths between them. I have to choose the 4 ways upwards (or the 5 ways to the right) of the 9 total (there are just 2 possibilities in the node, so when I choose 1 of them I'm done).

But if the lattice is in the 3D space, and I have 3 choices in each node, how can I solve the number of paths in this case? E.g k + m + n = 3 steps inwards, 4 steps upwards and 5 steps tho the right. And what if we have a lattice in any dimension?

Thanks!
Nick
 
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Are you sure about the 2D case? My reasoning is that, out of the n+m steps, we have to choose the m that are upwards, so the number of paths is C(n+m,m), which is more than n+m if m>1.

My approach leads to a natural extension to the formula for the number of paths in any number of dimensions. The answer will be a product of Combinations.
 
andrewkirk said:
Are you sure about the 2D case? My reasoning is that, out of the n+m steps, we have to choose the m that are upwards, so the number of paths is C(n+m,m), which is more than n+m if m>1.

My approach leads to a natural extension to the formula for the number of paths in any number of dimensions. The answer will be a product of Combinations.
Thank you. Yes, the number of paths should be C(n+m,m).

I will think more about the extension.

Tanks again!
 
nickek said:
I will think more about the extension.
Let me know how you go. I'm still working on calibrating my hints to steer a good path between too broad (a dead giveaway) and too narrow (not much help). Sometimes that challenge seems harder than solving the problem itself!
 

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