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Question

A,B,C are three persons among 7 persons who speak at a function. The number of ways in which it can be done if A speaks before B and B speaks before C is

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Solution

Number of ways in which 7 persons can speak in 7 places such that order of three particular persons A,B,C is fixed = 7!3!=840.

ALTERNATE SOLUTION:
Selection 3 places from 7 places can be done =7C3ways.
No of ways in which three particular persons A,B,C can speak in this function =7C3(as order is fixed)
Now, remaining 4 people can speak in 4 places in 4! ways
Required number of ways =7C3×4!=840

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