The way I see it, we can work it out like this:

Suppose an English team (Man U) is drawn for spot "A"... then the possibility of them facing another English team (Arsenal) is (1/7).  Arsenal's only chance to be paired with Man U is naturally in spot B.

Suppose next, an Spanish team (Sevilla) is drawn for spot "C".  The probability that they won't be paired with an Spanish team (Madrid) is 3/5.

For the remaining Spanish team the proability that they dont' get put against a Spanish teams is 2/3.

Multiplying these numbers gives:  6/105 = 2/35 or about 5.7%, not an impossibility, but a more-likely-than-not scenario!

Note:  Alternatively, we could calculate the probability by similarly placing Sevilla in slot C, then calculating that Madrid is placed in a favorable position: 4/5.  Similarly, the remaining team has a 1/2 chance of being placed in a favorable position.  This gives 2/35, the same result as above.

Both answers are correct, but I did it a different way using combinatorics.

It just dawned on me that there is only one way that draw could happen:
EE, Sx, Sy, Sz, x, y and z are the other 3 teams, giving 6 combinations.

Then it was a matter of finding out how many ways any quarter-final draw can be made. That gives 8!/4!2!2!2!2! = 105 possible random draws.

I adore Mathematicians ... I am still figuring out my basic linear equation. lolx..  **Thumbs up** @rycharde

Wow, welldone sir. I'm finding interesting   the calculative predictions of these probability deductions.. 

![DQmNSRf4McVosxb13FxAoMcB8rRoonJJiTfu7eaKHitndZJ.gif](https://steemitimages.com/DQmNSRf4McVosxb13FxAoMcB8rRoonJJiTfu7eaKHitndZJ/DQmNSRf4McVosxb13FxAoMcB8rRoonJJiTfu7eaKHitndZJ.gif)