If the letters of the word REGULATIONS be arranged in such a way that relative positions of the letters of…

The totals letters in the word REGULATIONS are $11$.

There are $11$ places corresponding to $11$ letters and in these $11$ places $R$ and $E$ can be arranged in ${}^{11}{P}_2$ ways. 

Now, Total possible cases where only $4$ letters are between $R$ and $E$ are $6$ as $\left(1, 6\right), \left(2, 7\right), \left(3, 8\right), \left(4, 9\right), \left(5, 10\right), \left(6, 11\right)$ where the numbers represent the place number and on reversing the order, we get total $12$ cases.

Hence, the required probability is $\frac{12}{{}^{11}{P}_2}$

$=12\times \frac{9!}{11!}$

$=\frac{6}{55}$