Probability of a quarter final combination There are three Spanish teams, two German, one Italian, one French, and one English. What is the probability that Germany plays itself? What is the probability that Spain plays itself? What is the intersection of this probability?
The first question I think is $1/28$ because ${8 \choose 2}$ = 28 and there is only one combination with Germany vs Germany so 1/28. Spain I think is 3/28 because ${3 \choose 2}$= 3 (three teams to choose from and only two teams needed). I would have thought that the intersection is where Spain play Spain and Germany play Germany. Apparently it's 3/35 but I don't know why. I tried using the $P(A \bigcup B)=P(A)+P(B)-P(A\bigcap B)$ but I don't know the union area.

We can list the teams in some order, say alphabetically.  There are seven ways we can match a team with the first team on the list.  This leaves six teams.  There are five ways to match a team with the first team remaining on the list.  This leaves four teams.  There are three ways to match a team with the first team remaining on the list.  The final two teams must play each other.  Hence, the number of possible quarterfinal pairings is 
$$7 \cdot 5 \cdot 3 \cdot 1$$

What is the probability that the two German teams play each other in the quarterfinal?

There is one way to match the German teams.  The remaining six teams can be matched in $5 \cdot 3 \cdot 1$ ways.  Hence, the probability that the German teams play each other is 
$$\frac{5 \cdot 3 \cdot 1}{7 \cdot 5 \cdot 3 \cdot 1} = \frac{1}{7}$$
which makes sense since given a German team, only one of the seven possible opposing teams is German.

What is the probability that two Spanish teams play each other in the quarterfinals?

As you observed, there are $\binom{3}{2} = 3$ ways to choose a pair of Spanish teams to play each other.  This leaves six teams, which can be matched in $5 \cdot 3 \cdot 1$ ways.  Hence, the number of quarterfinal schedules in which two of the Spanish teams play each other is 
$$\binom{3}{2} \cdot 5 \cdot 3 \cdot 1 = 3 \cdot 5 \cdot 3 \cdot 1$$
so the probability that two Spanish teams play each other in the quarterfinals is 
$$\frac{3 \cdot 5 \cdot 3 \cdot 1}{7 \cdot 5 \cdot 3 \cdot 1} = \frac{3}{7}$$

What is the probability that the two German teams play each other and two Spanish teams play each other in the quarterfinals?

There is one way to match the German teams and $\binom{3}{2} = 3$ ways to match two of the three Spanish teams.  This leaves four teams.  They can be matched in $3 \cdot 1$ ways.  Hence, the number of favorable schedules is 
$$\binom{3}{2} \cdot 3 \cdot 1 = 3 \cdot 3 \cdot 1$$
Hence, the desired probability is 
$$\frac{3 \cdot 3 \cdot 1}{7 \cdot 5 \cdot 3 \cdot 1} = \frac{3}{7 \cdot 5} = \frac{3}{35}$$
