Q:

I have five eggs to color for Easter. I can color them red, yellow, or blue. How many ways are there to do this?

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.

A:

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. 

A:

Consider a simpler problem. Suppose that you have just one egg. How many ways are there to color that egg using red, yellow or blue? The answer is: $3$, right?
Now, instead of one egg, you have two. How many ways can you color the second one? The answer is again $3$, right? 
But, recall that you had $3$ ways to color the first egg. Thus, for every possible color choice for the first egg, you can color the second one in $3$ ways. Therefore, the two eggs can be colored in $3 * 3$ ways.
So, the final answer is ...?

