FIFA Probability Problem I'm honestly unsure on how to go about this, so can anyone help me out please?
There are 32 teams competing in the upcoming World Cup. These 32 teams come
from ﬁve diﬀerent continents: Africa, Asia, Europe, North America, and South America.
Last December, FIFA (the orginization that governs soccer internationally) separated the 32
teams into four “pots”, each consisting of eight teams. Pot 1 contained the eight teams ranked
highest by FIFA, and consisted of four European teams and four South American teams. Pot
2 consisted of ﬁve African teams, two South American teams, and one European team. Pot
3 consisted of four Asian teams and four North American teams. Pot 4 consisted of eight
European teams.
In the World Cup, the initial stage consists of eight groups of four teams, with all groups
containing exactly one team from each pot described above. Suppose that FIFA creates the
groups via random selection, so that every possible outcome is equally likely. In actuality,
FIFA does not allow a single group to contain more than two European teams, or more than
one team from any other continent. If FIFA creates the groups via the random method
prescribed, what is the probability that at least one of these rules is broken?
 A: We can simplify the problem by eliminating pot 4 and changing the rule to: no team may contain more than 1 european. We will not take into account pot $4$ since it only contains europeans and won't affect the game once we modify the rule.Notice also that pot $3$ can be picked freely since it is the only one containing teams of north america or asia. So only the selections of pots 1 and 2 could incur in breaking any rules with the modified European rule.
So the probability a group division breaks no rules is the same as the probability the part of the groups corresponding to pots 1 and 2 don't have more than 1 european or more than 1 south american.
Pot 1 has 4 south americans and 4 europeans and pot 2 has 1 european, 2 south africans and 5 africans. The number of divisions not breaking any of these rules is the same as the number of bijections from the set $\{A_1,A_2,A_3,A_4,A_5,S_1,S_2,E_1\}$ To $\{S'_1,S'_2,S'_3,S'_4,E'_1,E'_2,E'_3,E'_4\}$ Where no $E$ is mapped to an $E'$ and no $S$ is mapped to an $S'$ How many of these are there? There are $4$ options for the image of $E_1$, and there are $4\cdot3$ ways to select the images of $S_1$ and $S_2$ Once we have selected those there are 5! ways to select where the E's are mapped. So in total there are 5760 ways to do it.
On the other hand there are $8!$ ways to select the group taking into account the first 2 pots . So the probability is $\frac{4*4*3*5!}{8!}=\frac{1}{7}$
This means the probability there is a broken rule is $\frac{6}{7}$
