The puzzlement is understandable. You want to count sequences, which brings permutations to mind. However, the most efficient way to solve this particular problem will use "combinations."
Put the $n-m$ working antennas in a row, leaving nice big gaps between them. If we want the bad antennas to be never next to each other, where can a bad antenna go? Into one of the gaps, or at either end. So there are $n-m+1$ potential places for bad antennas. We have to CHOOSE $m$ of these places. The number of ways to do this is $\binom{n-m+1}{m}$ (If $m$ is larger than $n-m+1$, then the job cannot be done at all. We can consider that the answer is still given by the above formula, if we define $\binom{x}{y}$ to be $0$ if $y>x$. Or else we can make the explicit restriction $m \le n-m+1$.)
It is (sadly) all too easy for a plausible sounding argument to be not correct. So one might check whether the above formula gives the right answer in "small" cases where one can count explicitly. And in addition one might see what happens for something like $n=10$, $m=3$, where you can take $7$ pennies, lay them out in a row, and try to think about where one can put $3$ dimes, so that no $2$ dimes are next to each other.