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In how many ways we can color $15$ eggs with colors red, blue, and green, when each egg must be colored with exactly two distinct colors.

My answer is :

(1) red and blue colored eggs are in the first box, and

(2) red and green colored eggs are in the second box, and

(3) blue and green colored eggs are in the third box.

So we have $3$ boxes. Any one or two of these boxes can be empty, because we can put all eggs in one box (that is color every egg with same combination of colors, so all are in one box).

Boxes are labeled and objects (eggs) are not.

So the answer is $$n+k-1 \choose k-1 $$ so $$ 15 + 3 -1 \choose 15-1$$ Is it correct?

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    I think you are right! I just think that you missed that last calculation. It should be ${15+3-1\choose 3-1}$ – 2017-02-14
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    Are the colourings oriented? Or is any red-greed egg indistinguishable from any other red-green egg? And are we allowed single-colour eggs? – 2017-02-14
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    No. I wrote every egg can be colored with two distinct colors only :) – 2017-02-14

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You've done a good job. But your final answer is off:

Using $n= 15,\; k=3$, and with the equivalence of $$\binom{n+k-1}{k-1} = \binom{n+k-1}{n},$$

your final binomial should be either

$$\binom{ 15+3-1}{3-1}\; \text{ or else its equivalent }\; \binom{15+3-1}{15}$$ both of which evaluate to $$\frac{17!}{2!15!} = \frac{17\cdot 16}{2} = 17\cdot 8 = 136$$

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    Can you dwell more on to how one gets $n+k-1 \choose k-1$? The boxes setup is clear but then how to come up with this result. Thanks in advance. – 2017-02-14
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    Thanks, so n is the number of elements, not boxes? right? – 2017-02-14
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    Take a look at the Wikipedia entry on Stars and Bars (Combinatorics), and specifically focus on the proof of Theorem 2. I can't seem to link to Wikipedia right now? Yes, $n$ is the number of eggs, and k is the number of boxes. – 2017-02-14
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    Margarita [Finally, I can link to the portion I'm referring to in the entry I refer to above!](https://en.wikipedia.org/wiki/Stars_and_bars_(combinatorics)#Theorem_two). – 2017-02-14
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    .........Thank you!:) – 2017-02-14
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    You're welcome, @Margarita. – 2017-02-14
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    Good answer. But I think the problem can be treated considering egg shapes as "elliptical" (answer showed above, because of symmetry) or "oval", where in this case k should be 6. :-) – 2017-02-15
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    @cgiovanardi Whatever shape you use to describe an egg in 3-D, the OP made it clear that every egg is colored with 2 distinct colors. Whether we have an egg with background green sparkled with red polka dots, or an egg covered in oval stripes, alternating red and green, they are to be considered indistinguishable. They each end up in the same box: the one labeled $\big(\color{green}{\mathrm{Green}}$ and $\color{red}{\mathrm{Red}}\big)$ – 2017-02-15
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    Anyway, @cgiovanardi, the only way I can make sense of your comment is that you were "joking"/having fun. It's a little bit cute, but if you didn't mean to be funny, then ... your comment is nonsense. – 2017-02-15
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    And indeed, [this egg, e.g.](https://www.colourbox.com/preview/17571851-greeting-card-colored-easter-egg.jpg) belongs in the box labeled $\big(\color{green}{\mathrm{Green}}$ and $\color{red}{\mathrm{Red}}\big)$ – 2017-02-15
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    @amWhy: For sure I don;t want to be rude, just interpreted the question in another way. And your egg brings another interpretation too (CW or CCW)! – 2017-02-15