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I started to solve this question and realised, that if I just add up all the possibilities, it is going to take a lot of time: Here is the complete question from the textbook:

Eggs that are all of the same size and shape are painted for Easter, and we are interested in how many eggs have which colour combination. Suppose each egg is painted with either a single colour or with two different colours. Compute the number of possibilities to:

a) Paint ten eggs if four colours are available,

b) Paint four eggs if ten colours are available.

I have started this way: all the eggs can be coloured in one colour, all the eggs can be coloured in two colours, 3 and 4. After that: all the eggs can be coloured in 2 colours and so on. But that seems to get a long journey. Is there any faster possibility solving this problem? I think, that those 2 problems are related. So solving the first is crucial for the second.

Thanks a lot.

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  • $\begingroup$ thank you Christian. I have edited it. My apologies $\endgroup$ – Eu2718 Oct 21 '18 at 22:55
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If there are $c$ actual colors available there are $c_*:=c+{c\choose2}$ distinguishable ways to color an egg with one or two colors. Given $n$ eggs we then have to find the number of nonnegative integer $c_*$-tuples $\bigl(x_k\bigr)_{1\leq k\leq c_*}$ satisfying $$x_1+x_2+\ldots+ x_{c_*}=n\ .$$ This is a stars and bars problem.

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  • $\begingroup$ thank you Christian $\endgroup$ – Eu2718 Oct 23 '18 at 23:53
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The ideas of solving these two problems are totally the same. Since A situation is too complexed, I only show the B situation. As shown in the description, a ball can be printed in a color or two colors. So, the colors available for balls are ten single color plus 45(10*9/2) combined colors, which is 55 in total.

If only one color is used in the way to color, the number of the way to color it is to pick one color from the 55 colors. So, it is 55 for only one used color.

If only two colors are used in the way to color, the number of the way to color it is to pick two colors from the 55 colors. And the way for the four balls can be 1+3, or 2+2, which numbers represents the times that a color is used. So there are 2*55*54/2+55*54/2 ways to color it.

If only three colors are used in the way to color, the number of the way to color it is to pick three colors from the 55 colors. And the way for the four balls can only be 1+1+2, which numbers represents the times that a color is used. So there are 3*55*54*53/(3*2) ways to color it.

Finally, if four colors are used in the way to color, the number of the way to color it is to pick three colors from the 55 colors. And the way for the four balls can only be 1+1+1+1, which numbers represents the times that a color is used. So there are 55*54*53*52/(4*3*2*1) ways to color it.

In sum, B situation has 427240 ways to color.

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  • $\begingroup$ thanks Xinye Bao $\endgroup$ – Eu2718 Oct 23 '18 at 23:54

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