What are the number of ways in which ten persons can speak, among whom A, B and C are to speak at a function, where A wants to speak before B and B wants to speak before C?

For a sequence of 10 people, the total unrestricted arrangements are 10!. Focusing on the relative order of A, B, and C, there are 3! possible sequences for these three people, but only one of those sequences satisfies the condition that A speaks before B and B speaks before C. Therefore, we divide the total permutations by 3! to get 10! / 6.

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