The book defines the probability of the negative binomial as:
$$P\{X=k\}={{k-1}\choose{n-1}} p^k (1-p)^{k-n}$$
but where does the ${k-1}\choose{n-1}$ come from? It's quite different to wikipedia's definition and I'm not sure what the probability is for here, the book doesn't say.
On
The random variable $X$ counts the number of trials until you reach $r$ ($n$ is a bad choice as a matter of notation) successes when the probability of success on an individual trial is $p$ independently of other trials.
Thus $X$ takes value $k$ (i.e. you need exactly $k$ trials to succeed $r$ times) if you succeed exactly $r-1$ times in the first $k-1$ trials which you can do in $$\dbinom{k-1}{r-1}p^{r-1}(1-p)^{k-1-(r-1)}$$ ways and succeed in the $k$-th trial with probability $p$ which results to $$P(X=k)=\dbinom{k-1}{r-1}p^{r-1}(1-p)^{k-1-(r-1)}\cdot p=\dbinom{k-1}{r-1}p^r(1-p)^{k-r}$$
Let it be that there is a probability of $p$ on succes by independent experiments. Then $X$ denotes the number of experiments needed to come to exactly $n$ successes.
We have $X=k$ if and only if the $k$-th experiment results in a succes and among the foregoing $k-1$ experiments there are exactly $n-1$ successes.
There are $\binom{k-1}{n-1}$ ways to choose $k-1$ experiments ('destined' to be successes) out of $n-1$.
All possibilities (including that the $k$-th experiment is a succes) have a probability of $p^n(1-p)^{k-n}$ to occur so that $$P(X=k)=\binom{k-1}{n-1}p^n(1-p)^{k-n}$$ (as Stef remarked in his comment).