I'm wondering how many ways you can paint n identical balls with k distinct colors.
I'm thinking of it as finding all possible sets of positive intergers solution to the equation
$$x_1+x_2+x_3+x_4+...+x_k=n$$
But that's where I'm stuck atm
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The stars and bars argument leads to the answer. https://en.wikipedia.org/wiki/Stars_and_bars_(combinatorics) – Sammy Black Nov 07 '13 at 22:06
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Must you use all k colors each time? Are all $x_i>0$? – DJohnM Nov 07 '13 at 22:17
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No, you just have to use at least 1 color – user102239 Nov 07 '13 at 22:22
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Also, see: http://math.stackexchange.com/questions/686/combinations-of-selecting-n-objects-with-k-different-types – Sammy Black Nov 07 '13 at 22:27
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So I read the 2nd theorem proof but was unable to understand the intuition. Could you explain it? – user102239 Nov 07 '13 at 22:30
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So the $x_i$ are non-negative, not positive... – DJohnM Nov 07 '13 at 23:47
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(This is Stars and Bars, just dissected a little.)
Take your $n$ identical balls and $k-1$ identical dividers, and arrange them in a row. This can be done in N ways:$$N=\frac{(n+k-1)!}{n! \cdot (k-1)!}$$bearing in mind that there are two sets of identical items.
Ues Color $\#1$ for the first section (up to the first divider), Color $\#2$ for the next section, and so on. This allows two or more dividers to come together, which deals with the "not this color" situation...
DJohnM
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