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Currently three Spanish teams are in the quarter-finals of the Championsleague. What are the odds that none of those teams play each other during the quarter finals?

All possible combinations is $8!$, is that correct?

But how many combinations are there in which Spanish teams do not play each other?

I tried something like this: Let's say the first team is Spanish, then it only has five options for non-Spanish teams etc... but it turns out it matters in what order you would pick the Spanish or non-Spanish teams, so now I am stuck. Any help would be appreciated.

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There are 3 possible ways they can meet: Sevilla vs Barcelona, Sevilla vs Real Madrid, Real Madrid vs Barcelona. These are disjoint, and the chance of any particular draw happening is $1/7$ (since any one team can play any of the other $7$, randomly. So the probability of two spanish teams meeting is $3/7$, and the probability of avoiding each other is $4/7$.

An alternative method is to consider this: the first spanish team is picked. There are $2/7$ ways for them to match one of the other two spanish teams. Suppose they don't. Then from the remaining $6$ teams, the chance of the two spanish teams matching is $1/5$, by the fact that any pairing is equally likely from the perspective of one of these remaining teams. So the probability of matching up is $$\frac27+\frac57\cdot\frac15=\frac37$$ as before.

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Total number of different pairwise combination of 8 teams = $\frac{8!}{2!2!2!2!4!}$ = 105

Total number of combination where 2 spanish team playing each other

= choosing 2 spanish teams out of three * different pairwise combination of remaining 6 teams

= $\binom{3}{2}$$\frac{6!}{2!2!2!3!}$= 45

Probability of 2 spanish teams playing each other = $\frac{45}{105}= \frac{3}{7}$

Probability of 2 spanish teams not playing each other= $\frac{4}{7}$

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