# In how many ways we can color $15$ eggs..

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.

(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?

• I think you are right! I just think that you missed that last calculation. It should be ${15+3-1\choose 3-1}$ – Arnaldo Feb 14 '17 at 19:41
• Are the colourings oriented? Or is any red-greed egg indistinguishable from any other red-green egg? And are we allowed single-colour eggs? – Joffan Feb 14 '17 at 19:43
• No. I wrote every egg can be colored with two distinct colors only :) – user296169 Feb 14 '17 at 19:45

Using $n= 15,\; k=3$, and with the equivalence of $$\binom{n+k-1}{k-1} = \binom{n+k-1}{n},$$
$$\binom{ 15+3-1}{3-1}\; \text{ or else its equivalent }\; \binom{15+3-1}{15}$$ both of which evaluate to $$\frac{17!}{2!15!} = \frac{17\cdot 16}{2} = 17\cdot 8 = 136$$
• 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. – rookie Feb 14 '17 at 20:06
• 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. – Namaste Feb 14 '17 at 20:16