Not sure if my thinking is correct on this problem. I have five eggs to color for Easter. I can color them red, yellow, or blue. How many ways are there to do this?
I was thinking 5 * 5 choose 3 since you would have 3 choices for color for each egg. Not sure if this would be correct though. Let me know what you think.
Best Answer
Let us assume that order does not matter, and that eggs of the same colour are indistinguishable. Then the problem is a standard Stars and Bars problem. The Wikipedia link gives a quite thorough explanation.
Briefly, we find the number of ways to distribute $5$ eggs (candies) between $3$ colours (kids). One or more kids may get nothing.
It is easier to think of the distribution as going as follows. We distribute $8=5+3$ candies among the $3$ kids, at least one to each kid, and then take away a candy from each kid. Or, if you want to be less cruel, we assign colours to $8$ eggs, with each colour being used on at least one egg, and then eat an egg of each colour.
Put down the $8$ candies like this $$\ast \qquad \ast \qquad\ast \qquad\ast \qquad\ast \qquad\ast \qquad\ast \qquad\ast$$ This determines $7$ intercandy gaps. Choose $2$ of these gaps to put a separator into, perhaps like this $$\ast \qquad \ast \quad|\quad\ast \qquad\ast \qquad\ast \qquad\ast \quad|\quad\ast \qquad\ast$$
This means Kid Red gets $2$ candies, Kid Yellow gets $4$, and Kid Blue gets $2$.
There are just as many ways to insert the two bars as there are to distribute the $8$ candies, at least one to each kid. Since there are $7$ intercandy gaps, there are $\binom{7}{2}$ ways to do the job.
Remark: For our case of $5$ and $3$, and indeed $n$ and $3$, the explicit enumeration of the answer by drhab is I think better. But the Stars and Bars technique comes up fairly often, so it is a good idea to get some exposure to it.