In last year's UEFA Champions League, for the first time, four teams from the same country, England (Arsenal, Chelsea, Liverpool, and Manchester United), all reached the quarter-finals. The eight quarter-finalists were paired through a blind draw, with the winners of each pair advancing to the semi-finals.

a) Some English fans would have preferred if the English teams avoided each other, as this would have opened up the possibility for all four teams to reach the semi-finals. What is the probability of such a pairing?

b) Other English fans would have preferred if two pairs of English teams were drawn against each other, as this would have guaranteed that two of their teams would advance to the semi-finals. What is the probability of such a pairing?

c) In the end, two English teams were drawn against each other, while the other two faced non-English opponents. What is the probability of such a pairing?
Solution. In the case of classical probability, the probability of an event can be calculated using the following relationship:

$$
p=\frac{\text { number of favorable cases }}{\text { total number of cases }} \text {. }
$$

Total number of cases:

First, we select 2 teams out of 8, then 2 teams out of 6, and finally 2 teams out of 4. These selections, since the order of selection is not significant, are combinations without repetition.

The same pairings can occur in multiple ways for the 4 pairs. Therefore, to get the total number of cases, we need to divide the number of combinations by the number of different orders of the same pairings.

$$
\text { total number of cases }=\frac{\binom{8}{2} \cdot\binom{6}{2} \cdot\binom{4}{2}}{4!}=105
$$

a) The English teams do not play against each other. We list the English teams, for example, in alphabetical order, and connect the non-English teams to them in every possible order. The number of these cases will be equal to the number of different orders of the non-English teams: 4! Based on this:

$$
p(A)=\frac{4!}{\frac{\binom{8}{2} \cdot\binom{6}{2} \cdot\binom{4}{2}}{4!}}=\frac{24}{105} \approx 0.2286
$$

b) The English teams play against each other. We select 2 teams out of 4, and divide by the number of different orders of the 2 pairs, and do the same for the non-English teams, as in the total number of cases: $\frac{\binom{4}{2}}{2} \cdot \frac{\binom{4}{2}}{2}=9$. Thus,

$$
p(B)=\frac{9}{\frac{\left(\begin{array}{l}
8 \\
2 \\
2
\end{array}\right) \cdot\binom{6}{2} \cdot\binom{4}{2}}{4!}} \approx 0.0857
$$

c) Two English teams play against each other, while the other two play against non-English teams. We select the two English teams from 4. For the other two teams, we select their opponents from the 4 non-English teams, but these can be paired with the English teams in two ways: $\binom{4}{2} \cdot\binom{4}{2} \cdot 2=72$. Therefore,

$$
p(C)=\frac{72}{\frac{\binom{8}{2} \cdot\binom{6}{2} \cdot\binom{4}{2}}{4!}} \approx 0.6857
$$

Note. In the third case, we can also calculate using the relationship $p(C)=1-p(A)-p(B)$.