If the letters of the word REGULATIONS be arranged at random, find the probability that there will be exactly four letters between the `R` and the`Edot`

Answer

`(11!)/(.^(9)P_(4)xx6! xx 2!)`

Explanation

Total number of ways of arranging 11 letters is (11)! The number of selection of 4 letters to be placed between R and E from remaining 9 is `.^(9)C_(4)`. These four letters can be permuted in 4! Ways. Now R and E can be interchanged in 2! Ways.
`underset("4 letter")ubrace((R....))underset(" 5 letter")ubrace(E")".....)`
Hence, the number of favorable ways is `.^(9)C_(4) xx 4! xx 6! xx 2!`
Therefore, the required probability is
`(11!)/(.^(9)P_(4) xx 6! xx2!)`