Question 1175182
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For the rest of 6 letters, C, D, E, F, G and H, there are 6! = 720 possible permutations.


For each of these 720 permutations, we have 7 positions to place the blocked AB or BA.


THEREFORE, the full number of possible arrangements, satisfying the posed conditions, is


    &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;6!*2*7 = 2*7! = 10080.      &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;<U>ANSWER</U>
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Solved.