101 uniforms all at least 0.001 apart

You draw $101$ independent uniform numbers on $(0,1)$. The probability that every pair of them is separated by more than $\tfrac1{1000}$ can be written in lowest terms as $\left(\tfrac ab\right)^c$. Find $a+b+c$.

$120$. For $n$ points with minimum gap $d$, removing a length-$d$ buffer after each of the first $n-1$ sorted points maps good configurations one-to-one onto unrestricted ones on an interval of length $1-(n-1)d$. So the probability is $(1-(n-1)d)^n=\left(\tfrac{9}{10}\right)^{101}$, and $9+10+101=120$.