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

If the letters of the word REGULATIONS bearranged at random, find the

1.	If the letters of the word REGULATIONS bearranged at random, find the probability that there will be exactly fourletters between the `R`and the`Edot`
Answer» Correct Answer - `(11!)/(.^(9)P_(4)xx6! xx 2!)` 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!)`

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

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