mathematics - Use two 2's, two 1's and two 8's to make the number 2018

This is another number formation question. You must use each of $2$, $1$ and $8$ exactly twice to make the number $2018$. The rules are

1. Allowed symbols: $+$, $-$, $\times$, $\div$, $($, $)$, $\sqrt{\quad}$, $!$. Arbitrary functions (such as the logarithm) are not allowed.


2. The ordering of the numbers are not important.


3. It is OK to use numbers as superscript (exponent or the power for the radical symbol).


4. Concatenation is allowed although using $(2+1)1$ to construct $31$ is not allowed.


5. Ceiling or flooring is not allowed. $211\div2=105.5$, not $105$ or $106$.



6. 
   

   The use of decimal point or scientific notation is not allowed.

   
   

   For example, $$(2+8)\times(2+1)!\times\sqrt{1+8}=180$$ is a valid construction, although this is not a solution because it does not equal $2018$.

   
   

Edit (after slvrbld posted his answer): Try to make $2108$ and $8102$ as well if you can.

(And of course, the more ways the better).

Second Edit I am thinking whether I can make all 4-digit numbers composed of 2, 0, 1 and 8 (numbers beginning with 0 do not count). I have solutions for some but not all of them. Once I find solution for all of them or find a well defined subset for which I have solutions, I will update it and make it a formal challenge here.

Answer

A possible solution:

$ 1\times2 + 8!/(12+8) $

Edit: more obscure approach:

$\displaystyle\binom{8\times8}{1\times1\times2} + 2 = \binom{64}{2} + 2 = 2018$