Derangements of 1-Factors Question

  • Thread starter Thread starter AXidenT
  • Start date Start date
Click For Summary

Homework Help Overview

The problem involves finding a formula for the number of ways to choose pairs of distinct 1-factors in the complete bipartite graph Kn,n that intersect in precisely i edges. The context includes concepts from combinatorics and graph theory, particularly focusing on derangements and matching in bipartite graphs.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss fixing one 1-factor and selecting intersecting edges, with questions about how to relate their ideas to the formula involving derangements. There is exploration of how to account for double counting and the correct interpretation of variables in the context of the problem.

Discussion Status

Some participants have provided guidance on the approach to take, including the use of derangements for the remaining edges. There is an ongoing exploration of how to formulate the problem mathematically, with some participants expressing uncertainty about the connections between their calculations and the sets involved.

Contextual Notes

Participants are navigating through the constraints of the problem, including the fixed integer i and the implications of choosing edges in the bipartite graph. There is mention of checking the formula against simple cases to validate the approach.

AXidenT
Messages
26
Reaction score
2

Homework Statement



"Let i be a fixed interger where 1≤i≤n-2. Given the graph Kn,n write down a formula for the number of ways you can choose a pair of distinct 1-factors which intersect in precisely i edges? Hint: fix one of the 1-factors, choose the i edges which will be common to both and then use derangements to choose the other n-i edges. Now how many ways can you choose the first 1-factor and when are you doubling counting?

Homework Equations



!n = n! nƩr=0 (-1)r/r!

The Attempt at a Solution



Following the hint, should I approach it by setting a 1 factor of A = {e1, e2, e3... en} And then set the intersection of A and the other 1-factor B as A\capB = {e1, e2... ei}

Problem is I don't know how to relate that idea to the formula we've been using - like what values to use for n and r? I would think n=(n-i), but that doesn't seem right. Am I completely off? Really having trouble with this one. :/ Thanks for any help!
 
Physics news on Phys.org
You want the remaining n-i edges of B all to mismatch the remaining n-i of A. If these n-i edges of A are the vertex pairs (u1, v1), (u2, v2) ... then you can form the edges for B by permuting the vi so that none of them stay the same. Sounds like a derangement to me.
 
Thank you for the reply! Could I calculate it like this?:

Number of Possible Matching for Kn,n * Number of intersecting edges in that matching * derangement of the remaining edges * 1/2 (to account for double counting from the first bit)

Where I think that would result in something like:

0.5 * n! * nCi * (n-i)! * n-iƩr=0 (-1)r/r!

It makes sense to me but I don't feel confident about how it relates to the sets. :S
 
AXidenT said:
Thank you for the reply! Could I calculate it like this?:

Number of Possible Matching for Kn,n * Number of intersecting edges in that matching * derangement of the remaining edges * 1/2 (to account for double counting from the first bit)

Where I think that would result in something like:

0.5 * n! * nCi * (n-i)! * n-iƩr=0 (-1)r/r!

It makes sense to me but I don't feel confident about how it relates to the sets. :S

That looks right. Have you checked it with some simple cases?
 
I just checked with n=2 and 3 and it seemed to work for me. Thank you! :D
 

Similar threads

  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
17
Views
2K
  • · Replies 7 ·
Replies
7
Views
3K
Replies
17
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 3 ·
Replies
3
Views
1K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 6 ·
Replies
6
Views
3K