Question 1194073
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if the letters A, B, C, D, E and F are randomly arranged, calculate the probability 
that the letters A and B will be next to each other?
Textbook answer is 1/3.
Can someone help me solve this problem? Thanks.
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<pre>
In all, there are 6 letters, A, B, C, D, E and F.   The numbers of all possible permutations is 6!.



IF the letters A and B form one block AB, then we have, actually, 5 objects to permute: 
4 letters C, D, E anf F plus one block AB.

5 objects can be arranged in 5! different ways.



If the letters A and B form one block  BA   (<U>in this order</U> !),  then another 5! different favorable permutations possible.


Thus the number of all favorable permutations with A and B next to each other is 2*5!.



Now the sought probability is the ratio of these two numbers

    P = {{{(2*5!)/6!}}} = {{{2/6}}} = {{{1/3}}}.    <U>ANSWER</U>
</pre>

Solved.