Among 10 persons, A, B, C are to speak at a function. The number of ways in which it can be don e if A wants to speak before B and B wants to speak before C is

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10!/24

9!/6

10!/6

none of these

answer is C.

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Detailed Solution

Places for A, B, C can be chosen in 10C3 ways. Remaining 7 persons can speak in 7! ways. Hence, the number of ways in which they can speak is 10C3×7! = 10!/6.

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