In how many ways can 14 tenth graders and 10 ninth graders be arranged in a line so that no two ninth graders may occupy consecutive positions? I have tried to solve it. But whatever way I came across, I ended up double counting. please help. thanks.

I only sum up the already present solution in the comments:

* 1st factor: $14!$ arrangements of $10$th graders
* 2nd factor: $\binom{15}{10}$ possible choices of "slots" to put exactly one $9$th grader
* 3rd factor: $10!$ arrangements of the $10$ $9$th graders within a given choice of slots



$$\mbox{Result: }14! \cdot \binom{15}{10} \cdot 10!$$