Identical balls arrangement in a circle Six identical yellow balls and four identical red balls are to be arranged in the circumference of a circle.  In how many ways it can be done?
 A: Out of 10 places we can choose 4 places for the red balls in $\binom {10}4$ =210 ways.  If they are not symmetric, they come in groups of $10$.  If the two pairs of red balls are opposite, they come in groups of 5.  For example, (1,2,6,7) is the same as (2,3,7,8), (3,4,8,9), (4,5,9,10) and (10,1,5,6) whereas (1,2,3,4) rotates in 10 different ways.  As there are 2 symmetric cases (accounting for 10 of the 210) there are 20 asymmetric cases (accounting for the other 200), for a total of 22.
Added:  to see the effect of rotational symmetry, consider two red and two yellow balls.  Before we allow for rotations, there are $\binom 42=6$ ways to position the red balls:  (1,2), (1,3), (1,4), (2,3), (2,4), and (3,4).  But when we rotate there are really only two choices:  the balls can be opposite or they can be adjacent.  If you look, (1,2), (2,3), (3,4), and (1,4) have them adjacent and (1,3) and (2,4) have them opposite.  So if there is no symmetry, we collapse 4 positions to one.  If there is twofold symmetry, we collapse 2 positions to one.  In the 6+4 case, we had cases of no symmetry, where we collapsed 10 positions to one and cases of twofold symmetry, where we collapsed 5 positions to one.
A: In general, the problem would correspond to counting binary necklaces of length $n$ (total of balls) with fixed density $m/n$, which is not trivial, see here (and try to understand Ross' answer).
At least, it should be easy to understand that $$\frac{{n \choose m}}{n}$$ is an lower bound, which should be quite precise if $n$ and $m$ are large. And, in particular, it's exact if $n$ and $m$ are relatively prime. 
This formula counts all possible placements of $m$ identical red balls in a $n$ urns in a ring, and divides by $n$ to account for rotational invariance. This division is not completly right, of course, because some rotations of some configurations can coincide.
A: If you arrange them, there are two possibilities: either it'll become the same in a cycle of five or ten. Let's first think about the first case, which should have 3 yellow and 2 red balls, and it'll be $\frac{5!}{3!2!}=10$. The second case, in the other hand, has 6 yellow and 4 red balls. Calculating $\frac{10!}{6!4!}$ would be 210, but we must subtract from the first case, and it's 200. Finally, we divide them, which is $\frac{10}{2}+\frac{200}{10}=22$.
