How many colours do we at least need so that we can ensure all 250 countries have different flags. One for FN standardized flag consists of three horizontal rectangular fields.
If we assume that the middle field not are allowed to have the same colour
as the top or bottom field, how many colours do we at least need to be sure
that we can ensure that all $250$ countries have different flags.
My attempt at a solution:
If the number of colours are $f$, then the top field can consist of
$f$ different colours, the middle field $f-1$ and the bottom field $f-1$. 
Hence, in total $f(f-1)^2$ colours.  We have that 
$f(f-1)^2 > (f-1)^3$ and because $5^3 < 250 < 6^3$ we can be sure
if $f -1 = 6$, i.e. $f=7$.
 A: You are correct that if there are $f$ colors available, then the number of different flags in which the middle field is not the same color as the top or bottom field is $f(f - 1)^2$.  Therefore, we require that $f(f - 1)^2 \geq 250$.  
\begin{array}{c | c}
f & f(f - 1)^2\\ \hline
1 & 0\\
2 & 2\\
3 & 12\\
4 & 36\\
5 & 80\\
6 & 150\\
7 & 252
\end{array}
From the table, we see that the smallest value of $f$ that satisfies the inequality is $7$.  
You attempted to estimate this value by using the observation that $f(f - 1)^2 \geq (f - 1)^3$.  You then wrote $5^3 < 250 < 6^3$, from which you concluded that we can be sure that if $f - 1 = 6$, then we have enough different flags, which led you to the correct conclusion, but your reasoning was incorrect.  Actually, $6^3 = 216 < 250 < 343 = 7^3$. If $f - 1 = 7$, then we may conclude that 
$$250 < 7^3 = (f - 1)^3 < f(f - 1)^2 = 8(8 - 1)^2 = 392$$
that is, we may conclude that $8$ is an upper bound on the number of colors we need.  However, the calculations above show that $f = 7$ is sufficient. 
A: So for the top and bottom we have $x$ colors. For the middle we can put all these colors but we must exclude on color in order to match requirements and not to go like $AAA$.
So sipmly: $x\cdot (x-1)\cdot x \geq 250$, $x\geq 7$.
