Probability of the Champions League quarter final draw featuring at least one all-English clash.

For a generalisation, suppose we have $2n$ teams being drawn into $n$ distinct fixtures, $k$ of the teams being English. To prevent an all-English matchup we must have $k\le n$ by the pigeonhole principle, then out of $(2n)!$ possible draws the number with no all-English matchups is the product of

• $\binom nk$ ways to choose which fixtures contain English teams
• $k!$ ways to assign the English teams once their fixtures have been chosen
• $2^k$ ways to decide for each English team whether they play at home first
• $(2n-k)!$ ways to place the remaining (non-English) teams

Hence the probability there is at one all-English matchup is $$1-\frac{2^k\binom nkk!(2n-k)!}{(2n)!}=1-2^k\cdot\frac{\binom nk}{\binom{2n}k}$$ For the Champions League situation of $n=k=4$, the probability works out to $1-16\cdot\frac1{70}=\frac{27}{35}$, in agreement with your result.