Question 1206447
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Six story books, A, B, C, D, E, and F were arranged on a shelf randomly. 
If A and B were not placed together, what is the probability that there was one book between A and B?
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(1)  In all, there are 6! = 6*5*4*3*2*1 = 720 different permutations of 6 books on the shelf.


(2)  Of them, there are 2*5! = 240 permutations, where the books A and B are together.


(3)  Hence, in the rest 720 - 240 = 480 permutations, the books A and B are not together.
     It is the number of all possible permutations under the condition "If A and B were not placed together".
     So, 480 is the denominator in the future fraction for the probability.


(4)  The number of all possible permutations of 6 letters with the block 
     of 3 letters ACB in 3 positions between #1 and #6 is 4! = 24.


(5)  The number of all possible permutations of 6 letters with the block 
     of 3 letters AXB in 3 positions between #1 and #6 with X= C or D or E or F 
     is 4*4! = 4*24 = 96.


(6)  The number of all possible permutations of 6 letters with the block 
     of 3 letters BXA in 3 positions between #1 and #6 with X= C or D or E or F 
     is also 4*4! = 4*24 = 96.


(7)  So, the numerator in the fraction for probability is  96+96 = 192.


(8)  Finally, the probability under the problem's question is  

        {{{192/480}}} = {{{16/40}}} = {{{4/10}}} = {{{2/5}}} = 0.4.    <U>ANSWER</U>
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Solved.