Theory of Numbers.

Notation: $\lfloor {x}\rfloor$ stands for the greatest integer no larger than a real number x.

By Legendre's formula, https://en.m.wikipedia.org/wiki/Legendre%27s_formula

$\displaystyle v_{11} (\frac{2018!}{1000!})$ = $\displaystyle\sum_{j=1}^{+\infty}\lfloor \frac{2018}{11^{j}} \rfloor$ $\displaystyle-\sum_{k=1}^{+\infty}\lfloor \frac{1000}{11^{k}} \rfloor$ = $\lfloor \frac{2018}{11}\rfloor$ + $\lfloor \frac{2018}{11^{2}}\rfloor$ + $\lfloor \frac{2018}{11^{3}}\rfloor$ - $\lfloor \frac{1000}{11}\rfloor$ - $\lfloor \frac{1000}{11^{2}}\rfloor$ = $\boxed{102}$

__ (note that $11^{3}=1331>1000$ and $11^{4}=14641>2018$ so in the two sums above there are two bunches of 0 in the end.)

Mathematica
We are searching for the prime factorization of the numerator and we are interested in the power that 11 is raised

Select[FactorInteger[2018!/1000!], #[[1]] == 11 &]

the above code gives {11,102} and 102 is the answer