Ten persons, amongst whom are A, B and C to speak at a function. The number of ways in which it can be done if A wants to speak before B and B wants to speak before C is
For A, B, C to speak in order of alphabets, 3 places out of 10 may be chosen first in 10C3 ways.
The remaining 7 persons can speak in 7! ways.
Hence, the number of ways in which all the 10 person can speak is 10C3.7! = 10!3! = 10!6