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There is a complicated mathematical way and there is a simple logical way to solve this problem.  Perhaps there exist some other ways of solving it.

Let N be the total number of different persons we have.    The number of persons (A, B, and C) we want them to be in one order is  a = 3 here.

1) Simple equipartition principle.
    Let us say we have arranged all the N persons in all possible permutations.  So that will be N! arrangements.  Let us say that in each arrangement the person on the left side speaks before the person on his/her right side.

All permutations of  a = 3 number of persons (A, B, C) : = 3! = 6
         A   B   C  ;   A  C  B  ;   B   C  A   ;  B A C  ;    C  A  B  ;   C  B  A

In the above, there is only one arrangement  A B C - the first one, which satisfies the condition of A speaking before B and B speaking before C.

The  N !  permutations of all N persons will be equally divided in to the above six groups like the following.  xx denotes some sequence of persons.
     xx A xx B xx C xx  ;  xx A xx C xx B xx  ;;  xx C  xx A  xx B
     xx B xx C xx A xx  ;  xx B xx A xx C xx  ;;  xx C  xx B xx A

So the number of arrangements we want are  N ! / a!  = 10! / 3! 

This can also be derived using permutations in another way.  Let the arrangements we want be in the following way:

   Let N = 10.  Let M = N - 3 = 7.  Let there be m number of persons speaking before A, n number of persons between A and B,  then p number of persons between B and C.  Finally,  after C there remain  (M  - m - n - p) number of persons to speak.

     ..m persons ..  A  ..n persons.. B .. p persons.. C .. (M - m - n - p) persons..

m varies from 0 to  M.    n varies from  0 to the remaining (M - m).
so p varies from  0 to  (M - m - n).

For each value of p  we have  (M - m - n - p) ! permutations of arrangements after C.  The number of ways of selecting p persons from available (M - m - n) number is:
           P(M - m - n,  p)        for p = 0 , 1, 2, ... , (M - m - n)
So total arrangements are :

\Sigma^{M-m-n}_{p=0} \ [ {}^{M-m-n}P_p \times (M-m-n-p)! ] \\\\=(M-n-m+1)!

The above number of ways is the way of arranging persons after B.  This is the number for each arrangement of  n persons between  A and B.  n varies from 0 to M-m.  The number of ways of selecting n persons is :
Let n' = M-m-n+1

\Sigma^{M-m}_{n=0}\ [ {}^{M-m}P_n * (M-m-n+1)! ]\\\\=\Sigma^{M-m+1}_{n'=1}\ [ (M-m)! * n' ]\\\\= (M-m)! * (M-m+1)(M-m+2)/2

Similarly: there are  P(M,  m) ways of selecting m  members before A and for each selection the above calculated number is the number of ways others follow A for speaking. now m varies from 0 to M.

let m' = M-m+1

\Sigma^{M}_{m=0}\ [ {}^{M}P_m * (M-m)! * (M-m+1)(M-m+2)/2] \\\\=\Sigma^{M}_{m=0}\ [(M! /2) (M-m+1)(M-m+2)]\\\\ =\frac{M!}{2} * \Sigma^{M+1}_{m'=1}\ [ (m'^2 + m') ]\\\\=\frac{M!}{2}*\frac{(M+1)(M+2)M+3)}{3}\\\\=\frac{(M+3)!}{3!}\\\\=N!/3!=10!/6
that is the answer...If you have N persons to speak and  n persons have to speak in one order, then the answer will be :   N ! / n ! 

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