Confused by combinatorical reasoning (n functional antennas, m defective problem) This is an example question and solution straight out of "A First Course in Probability" by Sheldon Ross, on page 6 (fyi: all that's covered till this point in the book is the basic and generalized principles of counting, counting orderings with repeated elements i.e. permutations of PEPPER, combinations).
Example 4c Consider a set of $n$ antennas of which $m$ are defective and $n-m$ are functional and assume that all of the defectives and all of the functionals are considered indistinguishable. How many linear orderings are there in which no two defectives are consecutive?
Solution Imagine that the $n-m$ functional antennas are lined up among themselves. Now, if no two defectives are to be consecutive, then the spaces between the functional antennas must each contain at most one defective antenna. That is, in the $n-m+1$ possible positions$-$represented in Figure 1.1 by carets$-$between the $n-m$ functional antennas, we must select $m$ of these in which to put the defective antennas. Hence, there are ${n-m+1}\choose{m}$ possible orderings in which there is at least one functional antenna between any two defective ones.
I saw this question, but I still don't get why the answer to this question has $n-m+1$. Can somebody please explain that?

Suppose there are a total of $7$ antennas, $2$ of which are bad. That means $n=7$, $m=2$, so the number of good antennas is $n-m=5$. We line up the five good ones, and ask where the bad ones can go:
__ G __ G __ G __ G __ G __
Each of the places where there's a line is an available spot for one (and no more than one!) bad antenna. Do you see how there are six available spots? That's $n-m+1$.
