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There are 100 balls numbered $n_{1}, n_{2}, n_{3}, . . ., n_{100}$ . They are arranged in all possible ways. How many arrangements would be there in which $n_{28}$ ball will always be before $n_{29}$ ball and the two of them will be adjacent to each other?

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99!.2!

99!/2!

99!

None of these

answer is C.

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Detailed Solution

Take the prefered two seats as one bunch, and arrange among them selves is $99!$ ways, and the specified two seats can be arranged in only one way

Hence, the total number of required ways is $99!$

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