Arranging Identical Chips in a Circle: Combinatorics Question Explained

[Mr Davis 97]

Homework Statement

In how many ways can four identical red chips and two identical white chips be arranged in a circle?

Homework Equations

The Attempt at a Solution

First, I calculated the number of different arrangements when the the chips are just in a line. This is $\displaystyle {6 \choose 2} = 15$. Next, I thought that since the chips are arranged in a circle, we have to divide by 6 to take into account the rotations that would essentially be considered the same. However, 15 is obviously not divisible by 6, so I am doing something wrong. What am I doing wrong?

 

[haruspex]

Homework Statement

In how many ways can four identical red chips and two identical white chips be arranged in a circle?

Homework Equations

The Attempt at a Solution

First, I calculated the number of different arrangements when the the chips are just in a line. This is $\displaystyle {6 \choose 2} = 15$. Next, I thought that since the chips are arranged in a circle, we have to divide by 6 to take into account the rotations that would essentially be considered the same. However, 15 is obviously not divisible by 6, so I am doing something wrong. What am I doing wrong?

Not all circles of 6 will produce 6 different linear sequences by starting at the 6 different points.
In general, you would also need to consider whether two circles that are mirror images should be counted as one or two, but in this very simple set up I don't think that arises.

 

[Mr Davis 97]

Not all circles of 6 will produce 6 different linear sequences by starting at the 6 different points.
In general, you would also need to consider whether two circles that are mirror images should be counted as one or two, but in this very simple set up I don't think that arises.

I looked at the problem a bit more and realized that there are only 3 arrangements, so this solves the problem. However, I still want to see how this is done with numbers. I'm not sure what you mean by the first part of your post. Do I need to rethink my initial approach, or do I need to somehow divide by 3 to get the correct answer?

 

[haruspex]

I looked at the problem a bit more and realized that there are only 3 arrangements, so this solves the problem. However, I still want to see how this is done with numbers. I'm not sure what you mean by the first part of your post. Do I need to rethink my initial approach, or do I need to somehow divide by 3 to get the correct answer?

There is no simple ratio that can be justified between the linear and circular counts.
If we start with the circle (RRWWWW) and generate linear sequences by taking the 6 possible starting points in it, we get 6 different sequences. Similarly with (RWRWWW).
If we start with the circle (RWWRWW), we only get three different linear sequences.
6+6+3=15.

 

[Stephen Tashi]

However, I still want to see how this is done with numbers.

Solving the general version of this problem is not simple, according to http://math.stackexchange.com/questions/163419/identical-balls-arrangement-in-a-circle/163429#163429 . To work deductively, we need to find some feature of this particular problem that makes it simpler than the general case.