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Q.

There are 100 balls numbered n1,n2,n3,...,n100. They are arranged in all possible ways. How many arrangements would be there in which n28 ball will always be before n29 ball and the two of them will be adjacent to each other?

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a

99!.2!

b

99!/2! 

c

99! 

d

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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