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.
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.