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If you have $11$ identical horses in how many ways can you paint 5 of them red 3 of them blue and 3 brown?

My intuition instantly tells me there is only one way of doing this. I mean if the horses were distinct I know there would be $11\choose{5,3,3}$ ways of painting them which is close to the answer given in the book I saw this in which was $\frac{1}{11} {11\choose{5,3,3}}$, but since they are identical i cant see how the answer isn't $1$. Did I misunderstand the problem ?

Here is the problem from the book itself.

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    $\begingroup$ Weird. In my view identical horses do not even exist (or at most one). They can be indistuingishable, but that is something else. Also the fraction $\frac1{11}$ makes no sense to me. $\endgroup$ – drhab Jul 17 '15 at 10:00
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    $\begingroup$ Yeah, that is why I hate applied or real world problems, you have to think of objects outside of pure mathematics. $\endgroup$ – alexgiorev Jul 17 '15 at 10:01
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    $\begingroup$ Makes one wonder how far identical (same property values) goes. A relaxed view would allow different coordinates, so you can distinguish them by the box you put them in. $\endgroup$ – mvw Jul 17 '15 at 10:02
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    $\begingroup$ Can we see the exact phrasing of the problem ? $\endgroup$ – Yves Daoust Jul 17 '15 at 10:06
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    $\begingroup$ Ah, carousel. dREaM (and Yves Daoust) had the right hunch. $\endgroup$ – mvw Jul 17 '15 at 10:09
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To understand the question you need to know something about the general behaviour of horses.

It is a well established fact that horses like to "horse around". What this tells us is that the horses are all positioned in a circular fashion, we need to find the ways to paint the horses, so that rotation of an arrangement counts as the same arrangement.

Now, if they where in a line the answer would be $\binom{11}{5,3,3}$. But each of these "linear" arrangements gives way to $11$ circular arrangements. Since closing the line and rotating it gives the $11$ arrangements.( to see this it is important to note $5$ and $3$ are both relatively prime to $11$).

Hence the answer is $\frac{1}{11}\binom{11}{5,3,3}$ as desired.

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  • $\begingroup$ Nice. Although there were shows where horses were put into two circles, running in opposite directions. $\endgroup$ – mvw Jul 17 '15 at 10:03
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    $\begingroup$ Yeah, but that can only be done when the number of horses is even. $\endgroup$ – Jorge Fernández Hidalgo Jul 17 '15 at 10:04
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    $\begingroup$ Interesting. From now on I am a mathematicien better skilled (i.e. with some knowledge about the behaviour of horses). $\endgroup$ – drhab Jul 17 '15 at 10:05
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    $\begingroup$ @dREaM, actually that's not true. The inner circle would be smaller, so would probably have one fewer horse; implying that it is more likely to be done with an odd total. $\endgroup$ – Ian MacDonald Jul 17 '15 at 17:04
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    $\begingroup$ @IanMacDonald : Only one fewer horse is unlikely -- A horse is about 8 feet long and the spacing is about 4 feet in those circles (sometimes less). Removing one horse would reduce the circumference by $\sim 12$ feet or the radius by $\sim 12/2\pi = 2$ feet, or less than the width of a horse. Unless these are ghost horses, then the interpenetration would be fine. $\endgroup$ – Eric Towers Jul 17 '15 at 19:59

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