How Are Circular Permutations of Indistinguishable Objects Calculated?

Click For Summary

Homework Help Overview

The discussion revolves around calculating the number of circular permutations of indistinguishable objects, specifically focusing on arrangements of m+n items where m items are alike and n items are also alike but of a different kind. The original poster attempts to apply a formula for circular arrangements but questions its validity based on specific examples.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss the original poster's approach of fixing one item and calculating arrangements based on that. They raise questions about the validity of the formula used and explore the implications of specific cases, such as when m = n = 2.

Discussion Status

Participants are actively engaging with the problem, questioning the assumptions behind the original formula and exploring the nature of arrangements. Hints have been provided to guide the original poster towards recognizing the limitations of their approach, but no consensus or resolution has been reached yet.

Contextual Notes

There is an acknowledgment of the difficulty in generalizing results for larger values of m and n, as well as the challenge of identifying repeating arrangements in circular permutations.

ritwik06
Messages
577
Reaction score
0

Homework Statement



Find the number of arrangements possible for arranging m+n things in a circular orientation, such that m things are alike and th other n things are also alike but of diffrent kind as from the first category.

Attempt:
I fix one thing. I am left with m+n-1
So the number of arrangements should be= [tex]\frac{(m+n-1)!}{(m-1)!n!}[/tex]
what is wrong with this approach?
 
Physics news on Phys.org
ritwik06 said:
Find the number of arrangements possible for arranging m+n things in a circular orientation, such that m things are alike and th other n things are also alike but of diffrent kind as from the first category.
…
I fix one thing. I am left with m+n-1
So the number of arrangements should be= [tex]\frac{(m+n-1)!}{(m-1)!n!}[/tex]
what is wrong with this approach?

Hi ritwik06! :smile:

Hint: if m = n = 2, there are only two possible arrangements … AABB and ABAB, but your formula gives 3!/2! = 3, because it includes ABBA.

Can you see why that's wrong, and how to deal with it? :wink:
 
tiny-tim said:
Hi ritwik06! :smile:

Hint: if m = n = 2, there are only two possible arrangements … AABB and ABAB, but your formula gives 3!/2! = 3, because it includes ABBA.

Can you see why that's wrong, and how to deal with it? :wink:

Yeah, Thats wrong.
So what should I do know? How can I check?? Is it only one arrangement that repeats itself?? Or Are there more? How can I find out? Making possible cases is easy when m,n are small but hen they are big its difficult. Help me please.
 
ritwik06 said:
Yeah, Thats wrong.
So what should I do know? How can I check?? Is it only one arrangement that repeats itself?? Or Are there more? How can I find out? Making possible cases is easy when m,n are small but hen they are big its difficult. Help me please.

Come on … think! :smile:

ABBA is wrong because … ? :wink:
 
tiny-tim said:
Come on … think! :smile:

ABBA is wrong because … ? :wink:

Its wrong because its the same as AABB...
But still, I want to generalise the result not by fixing m or n :eek:
 
Last edited:

Similar threads

Permutations with possibly repeating letters
  • · Replies 4 ·
Replies
4
Views
3K
Undergrad Understanding permutations and combinations in a coin toss experiment
  • · Replies 6 ·
Replies
6
Views
2K
Bishops and permutations on chessboard
  • · Replies 23 ·
Replies
23
Views
4K
High School Distributing N indistinguishable balls into M piles
  • · Replies 14 ·
Replies
14
Views
2K
Combinatorics Homework: Calculating Possible Combinations with Set A and B
  • · Replies 7 ·
Replies
7
Views
2K
Finding Permutations with Repetitions: n, x, a, b, c, d
  • · Replies 12 ·
Replies
12
Views
3K
What is the Equation for Combinations with Repetition in Combinatorics?
  • · Replies 5 ·
Replies
5
Views
2K
Possibly wrong permutation/combination question?
  • · Replies 3 ·
Replies
3
Views
2K
Number of valid ways to roll M dice with N sides
  • · Replies 8 ·
Replies
8
Views
3K
How Can Students Be Arranged into Groups with Set Sizes?
  • · Replies 12 ·
Replies
12
Views
3K