Number of ways in a 3D lattice

Click For Summary
The discussion focuses on calculating the number of paths in a 3D lattice, extending from a 2D case where the number of paths is determined by combinations. The initial calculation for 2D paths is debated, with clarification that the correct formula is C(n+m, m) for choosing upward steps. Participants explore how this concept can be generalized to higher dimensions, suggesting that the solution will involve a product of combinations. Resources such as videos on trinomial functions and path counting are shared for further understanding. The conversation emphasizes the complexity of extending these principles to multiple dimensions.
nickek
Messages
21
Reaction score
1
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
 
Mathematics news on Phys.org
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!
 
Good morning I have been refreshing my memory about Leibniz differentiation of integrals and found some useful videos from digital-university.org on YouTube. Although the audio quality is poor and the speaker proceeds a bit slowly, the explanations and processes are clear. However, it seems that one video in the Leibniz rule series is missing. While the videos are still present on YouTube, the referring website no longer exists but is preserved on the internet archive...

Similar threads

  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 7 ·
Replies
7
Views
3K
  • · Replies 5 ·
Replies
5
Views
1K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K
Replies
2
Views
2K
  • · Replies 8 ·
Replies
8
Views
2K
Replies
1
Views
3K
Replies
1
Views
6K
  • · Replies 5 ·
Replies
5
Views
2K