In how many ways can 4 identical red balls and two identical white balls be arranged in a circle? This is an elementary problem, but many tries have not yet yielded results. I tried by taking the total arrangements as (n-1)! because it is a circle but then the answer is fractional. Could someone explain how I might go about with it?
Many thanks.
On
Yes, @mapierce271 is right.
Let us assume that the first ball is white. (First ball we look at, since a circle has no 'first' point).
I would start by seeing the number of balls between the 2 white balls:
a) 0 - Yes, it is possible. WWRRRR
b) 1 - This, too, can be done. WRWRRR
c) 2 - Again. WRRWRR
d) 3 - This would lead to WRRRWR, which is a cycled arrangement of b)
e) 4 - This would be WRRRRW, which is another way of writing a)
So, only a), b) and c) are unique and correct.
The question is easy since there are only 2 white balls. It gets complicated with more than 3.
I think that you are over-thinking it. Since the balls are identical, there are only $3$ arrangements: $$ \dots WWRRRR \dots\quad \dots WRWRRR \dots\quad \dots WRRWRR \dots $$ where $\dots$ means that the circle wraps around.