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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?

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Are rotated arrangements considered equivalent to each other? That is, do you consider 6 yellows followed by 4 reds to be one arrangement or 10 arrangements? –  copper.hat Jun 26 '12 at 18:01
    
@copper.hat : balls are identical so one arrangement –  Aizen Jun 26 '12 at 18:07
    
@Aizen: you've answered copper.hat's question (and that's the reasonable answer) but the reason you give is not right, that the balls are identical has nothing to do with it. –  leonbloy Jun 26 '12 at 18:41
    
Does your circle have an orientation? Or is it like a necklace with ten beads that we can choose to wear upside down. In other words are for example RRRYRYYYYY and YYYYYRYRRR (when bent to form a circle) the same. Or yet in other words, can we distinguish mirror images from each other? –  Jyrki Lahtonen Jun 26 '12 at 18:44

2 Answers 2

up vote 5 down vote accepted

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.

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can you please elaborate your solution.I couldn't get how you got 22.If you can make the result generalized for like n identical yellow and m identical red balls it would be helpful. –  Aizen Jun 26 '12 at 18:28
    
@Aizen: did you try with a smaller example, say 2 red balls and 4 yellow balls? what result did you get? –  leonbloy Jun 26 '12 at 19:11

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.

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