In how many ways $A$ speaks before $B$ and $B$ speaks before $C$ $10$ persons has to give a speech among which three are $A$, $B$ and $C$. In how many ways can they give speech so that $A$ speaks before $B$ and $B$ speaks before $C$.
I have taken the fixed speech order of $A$,$B$, $C$  as $$*A*B*C*$$
where the stars represent remaining $7$ persons can be accommodate.
That means i have to find number of non negative integral solutions of $$x_1+x_2+x_3+x_4=7$$.
But i have no idea how can arrangements be done in a particular star.
 A: Your idea will work. Using Stars and Bars we find that there are $\binom{10}{7}$ ways to decide the positions to be occupied by "others." Once these positions are decided, they can be filled in $7!$ ways. We get a total of $\binom{10}{7}7!$, which can be simplified greatly.
There are many other approaches.  Maybe the shortest is to note that the $10$ people can be permuted in $10!$ ways.  By symmetry, the fraction of these in which A, B, C are in the right order is $\frac{1}{3!}$, for a total of $\dfrac{10!}{3!}$.
A: The simplest way is to think of all the ways of ordering the speeches.  By symmetry, $\frac 1{3!}$ of these will have A before B before C because you can group them in batches that only reorder A,B,C.  I am assuming the other speakers are distinguishable.
A: I think you're basically done. If we have 10 spaces for each of 10 speakers, and we place the other 7 speakers first, then we will be left with 3 spaces no matter how we arrange the other 7. And in these 3 spaces it will be no issue to simply place them in the order A, B, C. Therefore the real issue of the problem becomes arranging the other 7 speakers. They can be arranged in $10P7$ ways, and so after this the other 3 speakers can be arranged in the order you wish. And so $10P7$ is the answer.
