How Are Circular Permutations of Indistinguishable Objects Calculated?

[ritwik06]

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= \frac{(m+n-1)!}{(m-1)!n!}
what is wrong with this approach?

 

[tiny-tim]

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= \frac{(m+n-1)!}{(m-1)!n!}
what is wrong with this approach?

Hi ritwik06!

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?

 

[ritwik06]

Hi ritwik06!

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?

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.

 

[tiny-tim]

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!

ABBA is wrong because … ?

 

[ritwik06]

Come on … think!

ABBA is wrong because … ?

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