Hint:
If we think rotated rectangles are different: so
Red Green Blue Black
Black Blue Red Green
is different from
Green Red Blue Black
Black blue Green Red
There are 12 choices for the first column. How many choices are there for the second column? For the third?
If you worry about rotations, it becomes harder. You might also count rectangles that can be made to match by reassigning colors the same. Then you can arbitrarily start with Red Green in the first column, but it is also hard.
Added in response to comment: in the case where you count rotations as the same, we have double counted all the rectangles except the symmetric ones, so we just need to count those. The last half of the columns are specified when you pick the first half. So again you have 12 choices for the first column and 7 for the second. As dtldarek pointed out, you can always invert the second column to make the third and the first to get the fourth. So there are 84 symmetric cases. The total with rotations the same is then $(12\cdot7^3-84)/2+84=2100$.
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.
Best Answer
It is a interpretation problem, my advice is to go to the simplest. Imagine the subject allow you to paint each wall with different color, the how many walls have the room ? some have 6 or more walls. It does not seems logic. If it is not precised assumed it is not asked.
The simplest is considering you can paint the rooms with only one color each. That way, you have 2 choice for the first, times 2 choice for the second etc. i.e. $$2\times2\times2...\times2=2^{14}$$
Have a nice day.