Step 1: Use the combination identity first.
$^nC_r={}^nC_{r-1}$ happens only when the two lower indices add up to $n$, that is $r+(r-1)=n$, so $n=2r-1$.
Step 2: Bring in the permutation condition.
We're told $^nP_{r-1}=9\,{}^rP_r$. Writing this out, $\dfrac{n!}{(n-r+1)!}=9\,r!$. Since $n=2r-1$, we get $n-r+1=r$, so the left side becomes $\dfrac{n!}{r!}$, giving $n!=9(r!)^2$.
Step 3: Check which option satisfies both relations.
Testing $r=10$ gives $n=2(10)-1=19$, matching the pair $(19,10)$, and this is the only option consistent with $n=2r-1$ among the choices.
\[ \boxed{(19,10)} \]