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To save your time, I simplify the question to something like this:

There are $18$ people, $4$ of which are teachers. All of them($18$) are going to stand in a row. In how many ways can they be arranged such that no teachers stand next to each other?

And the answer for this question is around $2.32\times10^{10}$. I have no idea how to get to the answer. Can anyone point me in the right direction?

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So, you have $14$ ordinary people and $4$ teachers, and told to arrange in such a way, that no teacher stand together. First place the ordinary $14$ people. So, there are $15$ places to allocate the teachers. The diagram below illustrates this. $$-|-|-|-|-|-|-|-|-|-|-|-|-|-|-,$$ where $|$ denote the $14$ ordinary people, and $-$ denote the spaces between them. This can be done in $14!$ ways.

Now, if you place the $4$ teachers in any of these $15$ places. they can never be adjacent. So, the answer should be $$14!\times\binom {15}4,$$ and also, if the order of those teachers is relevant, then, the answer should be $$14!\times4!\times\binom{15}4.$$

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  • $\begingroup$ Yeah I used a similar approach, however, I could not get to the answer... $\endgroup$ – Benson Apr 9 '16 at 14:51
  • $\begingroup$ Guess the answer might be wrong then... $\endgroup$ – Benson Apr 9 '16 at 15:00

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