UEFA Champions League quarterfinals 2018 draw - pairing of same country teams The same question - asked by a first-time user yesterday - was deleted after downvoting for lack of genuine effort. But I still would like to know the answer...
The issue is that in the lottery to establish the quarterfinal draw of this football tournament none of the 3 teams from Spain (BCN, RM, SVA) were paired among themselves. Instead, they were paired with ROM, JUV, BYN, respectively:

In addition, the two English teams were paired up (LIV - MC).
I had the impression that the question could conceivably have originated in a discussion among English fans, understandably disappointed about the fact that only 1 team from England can advance to the next stage in the competition, while all teams from Spain can theoretically advance. Just a framing story...

The question was what is the probability of all teams from Spain avoiding each other, and at the same time, the 2 teams from England having to face each other?


About showing some effort:
 - The sample space of outcomes:
The first ball drawn can correspond to any one of the $8$ teams, and whichever this happens to be, it can be paired with one of the other $7$ teams. At that point a team will be chosen among the remaining $6$ teams, and paired with $5$ possible teams. The third pair will be chosen by first drawing among $4$ teams with $3$ possible opponents. The last pair is at that point determined (disregarding order - i.e. field advantage). So
$$8 \times 7 + 6 \times 5 + 4 \times 3 = 98.$$

Side notes:
Why is the above calculation wrong? Thank you @RossMillikan:

Three things. You should be multiplying the terms because for each first pair you can choose any two teams next for the second pair. You should be dividing each pairing (including the last 2∗1) by $2$ because you don't care which order the teams are picked for a pair. You should be dividing by $4!=24$ because you don't care what order the pairs are picked. AB,CD,EF,GH is the same pairing as HG,DC,AB,EF. This is what gets to $105.$ 

To see this implemented in a computer simulation, see this.

I found the formula for the number of possible pairings of $n$ teams here:
$$\frac{{n \choose 2}{n-2 \choose 2}\dots{2 \choose 2}}{\left(\frac{n}{2}\right)!}=\frac{n!}{2^{n/2} (n/2)!}=\frac{8!}{2^4 4!}=105$$
which makes sense from the derivation, but it is at odds with the comment below:

Imagine that $8$ teams participate in a round robin tournament (i.e. each team play all other teams once). The total number of matches will be the total number of possible pairings. The first team has $7$ matches, the second team has $6$ (match with the first team already counted) and so forth. The total number is $7+6+5+4+3+2+1=28.$

Is it $105$ or $28$?
Again, thanks to @RossMillikan, the comment that clears this issue is:

$28$ and $105$ are counting two different things. $28$ counts the number of games in a round robin of $8$ teams. $105$ counts the number of ways to break $8$ teams into four matches of two teams each.


Thinking of the formula for number of pairs, there is a very intuitive way of seeing that is correct. If a vector of elements corresponding to the different teams $\{1,2,3,4,5,6,7,8\}$ is read as "team $1$ is going to face team $2$; team $3$ will face team $4$, etc", all we have to do is scramble this vector into all its possible permutations $8!$ In this way, $\{1,5,3,4,2,6,7,8\}$ will now pair team $1$ with team $5,$ for example.
After this all we need to do is follow the comment by RM, and divide $8!$ by first, $4!$ because the order of the pairs does not matter, i.e. $\{1,2,3,4,5,6,7,8\}= \{3,4,1,2,5,6,7,8\},$ for example. This would give the total possible of draws including the field advantage in the first leg: $8!/4!=1,680.$
Further, since we don't care right now about whether BCN plays RMA first in Barcelona and later in Rome or vice versa, and there are $4$ pairs, we still have to divided by $2^4$ to get $\frac{8!}{2^4 \times 4!}=105.$ 

- "Favorable" outcomes:
As for favorable outcomes, there are 3 teams from Spain and 3 non-English teams to be paired up with. Hence, $3\times 3 =9$ possible pairings. The fact that the two English teams are paired up in this question can only happen in $1$ possible way LIV - MC.
Side notes:
This is wrong, after the answer by Remy. I see why: It would allow RM - ROM and BCN - ROM as possible, while ROM can only be paired with either RM or BCN, for example. So there is overcounting. The right way to look at it is: For the first team from Spain, there are $3$ non-Spanish teams available; for the second team from Spain, there are $2$ non-Spanish team left; for the last team from Spain there is only one pairing available. Hence, it is $3\times 2 \times 1 =6.$

So the $$\require{cancel}\cancel{\Pr(\text{2018 outcome}) = 9/98 \approx 9 \text{ percent}}.$$
Side notes:
So it's either
$$\require{cancel}\cancel{\frac{6}{28}=21\text{ percent}}$$ or $$\frac{6}{105}=5.7\text{ percent}$$
... the latter with the ring of truth in it...
 A: Put Liverpool anywhere. Then the probability that they are paired with Manchester City is $\frac{1}{7}$. 
Then put down a Spanish team in one of the remaining $6$ slots. The probability that they are paired with a non-Spanish team is $\frac{3}{5}$.
Then put down one of the remaining $2$ Spanish teams. The probability that they are paired with a non-Spanish team is $\frac{2}{3}$.
Then the remaining two teams (one Spanish and one non-Spanish) are guaranteed to be matched together.
All together, we get $$\frac{1}{7}\cdot\frac{3}{5}\cdot\frac{2}{3}\approx0.057$$

Looks like your edited sample space count is correct. For the event space, you over-counted. As you correctly stated, there is only one way for the English teams to play each other. As for the other six teams, consider lining up the $3$ Spanish teams as follows:
$S_1$
$S_2$
$S_3$
Then there are $3!=6$ ways to arrange the $3$ non-Spanish teams, which I will denote as $N_1,N_2,N_3$. Our event space is small enough that we can write down all the possibilities. 
$S_1N_1,S_2N_2,S_3N_3$
$S_1N_1,S_2N_3,S_3N_2$
$S_1N_2,S_2N_3,S_3N_1$
$S_1N_2,S_2N_1,S_3N_3$
$S_1N_3,S_2N_2,S_3N_1$
$S_1N_3,S_2N_1,S_3N_2$
