Why does negative binomial random variable uses ${n-1 \choose r-1}$ instead of ${n \choose r}$ as coefficient? negative binomial random variable works like this, but can you explain why we use ${n-1 \choose r-1}$ instead of  ${n \choose r}$?
$$P(X=x) = {x-1\choose r-1}(1-p)^{x-r}p^r$$
 A: Basic reason the use of ${n-1} \choose {r-1} $  lies in the description of the situation where negative binomial occurs.We want rth success in nth trial.Hence,$r-1$ success in $n-1$ trial and nth trial being a success.Hence $n-1 \choose r-1 $.
Now you can redefine the situation in which $n \choose r$ makes sense but try doing it,it is not very easy to perceive.
EDIT:As asked in comments,little elaboration.
Lets start from scratch:
We want first $n-1$ trials result in $r-1$ successes and $(n-1)-(r-1)=n-r$ failures.Any of $r-1$ out of $n-1$ trials can be success,and remaining $n-r$ should be be failures (note that choosing $r-1$ success also fixes $n-r$ failures.Hence ${n-1 \choose r-1} \times p^{r-1} \times {(1-p)}^{n-r}$ . Now the last trial must be success,probability of which is $p$,hence we multiply by $p$ and get required answer.
Hope this helps.
A: Dunno, maybe for the sum of independent NB$(r,p)$ and NB$(s,p)$ to be NB$(r+s,p)$. 
Slightly less convincing: for NB$(r,p)$ to be geometric when $r=1$.
