probability and combinations with the word REGULATIONS 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?
 A: 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 $\frac{6}{\binom{11}{2}}$. 
A: 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
A: 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.

A: 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
