Three Color Flag

This is similar to the number of ways you can arrange the letters in the word AAABBBCCC, for which the solution is:

$\frac{9!}{3!3!3!} = 1680$

The reason they are essentially the same question, is that if each of the letters corresponds to one of the colors (e.g. A=green, B=white, C=yellow) and each of the positions of the letters corresponds to the one of the squares in the flag, then we have a 1:1 mapping of the solutions of the flag problem to the rearrangements of AAABBBCCC.

Let's start using the Yellow color, to paint the flag. The yellow color should appear in 3 small rectangles. These rectangles can be any of the 9 rectangles. How there are no differences between the rectangles, the number of ways that we can choose 3 rectangles between 9 without repetitions is:

$\binom{9}{3}=\frac{9!}{3!\left ( 9-3 \right )!}=84$

Now let's use the Green color.

How there are left six rectangles to paint, the number of ways that we can choose 3 rectangles from it is:

$\binom{6}{3}=\frac{6!}{3!\left ( 6-3 \right )!}=20$

Now there are left 3 rectangles left. These rectangle will recieve the last colour (White). Not matter the way that we paint the flag, always will be left 3 rectangles. Then the number of ways that we can paint this flag is: $84\times20=1680$ .