The first $12$ letters of English alphabet are written down at random in a row. The probability, that there are exactly $4$ letters between A and B, is 

A and B can occupy the positions,  $(1, 6), (2, 7),(3, 8), (4, 9), (5, 10), (6, 11), (7, 12)$

in 7 ways and A,B can be interchanged in 2 ways .

Other 10 people can be arranged in  10! ways

$\therefore$ $$\begin{gathered} n\left(E\right)=2\times 7\times 10!,n\left(S\right)=12! \\ P\left(E\right)=\frac{2\times 7\times 10!}{12!}=\frac{7}{66} \end{gathered}$$