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Q.If the letters of the word PROBABILITY are arranged at random, the probability that no two B’s are together is

a
8!×911!\dfrac{8! \times 9}{11!}
b
8!×(92)11!/2!\dfrac{8! \times \binom{9}{2}}{11! / 2!}
c
9!×(92)11!\dfrac{9! \times \binom{9}{2}}{11!}
d
8!×911!/2!\dfrac{8! \times 9}{11! / 2!}

Correct Answer: Option A

The correct solution involves applying the fundamental concept to derive the final value step by step...

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