Choose between Combinations & Permutations What is the most natural way to choose either permutations or combinations when encountering a particular combinatorial problem?

For example, in the following problem:

Given $n$ antennas of which $m$ are defective and  $n-m$ are functional, how many linear 'orderings' are there in which no two defectives are consecutive? 

What seems to be the natural way of doing this (P or C)?
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 $y&gt;x$.  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.
lets change the question to daily situation for understanding.
consider n students in which m are boys so we have n-m girls.how many linear orders are possible if no two boys sit consecutive?
[In our example consider 7 students in which there are 2 boys then n-m=5 so we have 5 girls]
GGGGG arrangement of n-m girls
since no boy should sit consecutively they can sit beside each girl
BGBGBGBGBGB [so there are 6 places for boys 5+1=n-m+1]
so we need m positions out of n-m+1 [we need 2 position out of 6 positions available]
possible combiantions are( n-m+1Cm)