If the letters of the word 'REGULATIONS' are arranged at random, then the probability that there will be exactly 4 letters between R and E is

Total number of ways of arranging 11 letters = (11)! ---------- (1) 

Selection of 4letters to be placed between R and E from the remaining 9 ${=}^9{C}_4$ 

These 4letters can be permuted in 4! ways. Now permutation of 

$R\text{ and }E\text{ can be interchanged in }2!\text{ ways. }$

$\therefore \text{ Number of favourable ways }{=}^9{C}_{4}6!2!------\left(2\right)$

$\text{ Hence, required probability }=\frac{11!}{{ }^9{C}_{4}6!2!}$