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If the letters of the word REGULATIONS are arranged at random,what is the probability that there will be exactly 4 letters between R and E? The answer in my book is given as 11!/(9C4 x 4! x6!x2!) .Shouldn't the answer be upside down because 11!=total number of arrangements?

ayush
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4 Answers4

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There are $\binom{11}{2}$ equally likely ways to choose the two positions that will be reserved for the letters R and E. (Note this does not count where R and E as individual letters go, we are just putting reserved signs on the two positions.)

If the two positions have a gap of $4$ between them, there are $6$ places where the leftmost position can be, and then the other position is determined. It follows that our probability is $\frac{6}{\binom{11}{2}}$.

André Nicolas
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  • Could you plz respond to my edited question?The answer in my book is given as 11!/(9C4 x 4! x6!x2!) how to reach this answer or is it a misprint in the book? – ayush Nov 23 '13 at 09:05
  • The answer given is wrong. If you turn it upside down it will be OK. Turned upside down, it is an awkward but correct way of doing it. – André Nicolas Nov 23 '13 at 14:08
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Ans: 6/55

Totally there are 11 letters.

R,E can occupy 6 positions like (1,6) (2,7) (3,8) (4,9) (5,10) (6,11). In this R and E can interchange among themselves.

In remaining positions the 9 other letters can arrange themselves in 9! ways. So total ways where R and E have 4 letters between them = 6 * 2 * 9!

Required probability = (6 * 2 * 9!)/(11!) = 12/110 = 6/55

Karthikeyan Chockalingam
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You could take $\dfrac{{\binom{9}{4}}\times4!\times2\times6!} {11!}$ and get 6/55 as your answer.

  • $\binom{9}{4}$$\times4!$ because you have to choose the four numbers apart from R and E that should be tucked in between R and E, after which you should account for the possible arrangements of those four numbers.
  • $2$ because you have to account for R-E and E-R arrangements
  • $6!$ because you take R- insert 4 letters - E as a block, this block plus the 5 other remaining numbers can be arranged in 6! ways
  • 11! because that shows the possible arrangements of all the eleven letters.
Meera Unni
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Here R E can occupy (1,6),(2,7),(3,8),(4,9),(5,10),(6,11) and vice versa is also possible .so the number of ways to fill this is 6*2 ways. Then remaining 9 letters can fill in 9! ways. So the probability will be 6*2*9!/11! ( Since there are 11 letters in total. By solving this we get 6/55

And:6/55

Vijay Suriya
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  • The Question you are responding to is four and a half years old, and already had Accepted and upvoted Answers a long time before. While you reach the same solution $6/55$ as the three earlier Answers, you do not highlight any new information or respond to the OP's concern about the "answer in my book". You should perhaps strive to contribute Answers to Questions for which you can contribute something new. Good luck! – hardmath Jun 05 '18 at 22:05