Let $P(x) = 1-x+x^2-x^3+\dots+x^{18}-x^{19}$ and $Q(x)=P(x-1)$. What is the coefficient of $x^2$ in polynomial $Q$?

$ 
\textbf{(A)}\ 840
\qquad\textbf{(B)}\ 816
\qquad\textbf{(C)}\ 969
\qquad\textbf{(D)}\ 1020
\qquad\textbf{(E)}\ 1140
$
1. First, we start with the polynomial \( P(x) = 1 - x + x^2 - x^3 + \dots + x^{18} - x^{19} \). This is a finite geometric series with the first term \( a = 1 \) and common ratio \( r = -x \).

2. The sum of a finite geometric series can be expressed as:
   \[
   P(x) = \sum_{k=0}^{19} (-x)^k = \frac{1 - (-x)^{20}}{1 - (-x)} = \frac{1 - x^{20}}{1 + x}
   \]

3. Next, we need to find \( Q(x) = P(x-1) \). Substituting \( x-1 \) into \( P(x) \), we get:
   \[
   Q(x) = P(x-1) = \frac{1 - (x-1)^{20}}{1 + (x-1)} = \frac{1 - (x-1)^{20}}{x}
   \]

4. To find the coefficient of \( x^2 \) in \( Q(x) \), we need to expand \( 1 - (x-1)^{20} \) and identify the term that will contribute to \( x^2 \) when divided by \( x \).

5. Using the binomial theorem, we expand \( (x-1)^{20} \):
   \[
   (x-1)^{20} = \sum_{k=0}^{20} \binom{20}{k} x^k (-1)^{20-k}
   \]

6. We are interested in the \( x^3 \) term in the expansion of \( -(x-1)^{20} \) because when divided by \( x \), it will contribute to the \( x^2 \) term in \( Q(x) \).

7. The \( x^3 \) term in the expansion of \( -(x-1)^{20} \) is:
   \[
   -\binom{20}{3} x^3 (-1)^{17} = -\binom{20}{3} x^3 (-1) = \binom{20}{3} x^3
   \]

8. Dividing this term by \( x \) gives:
   \[
   \binom{20}{3} x^2
   \]

9. Therefore, the coefficient of \( x^2 \) in \( Q(x) \) is \( \binom{20}{3} \).

10. Calculating \( \binom{20}{3} \):
    \[
    \binom{20}{3} = \frac{20!}{3!(20-3)!} = \frac{20 \times 19 \times 18}{3 \times 2 \times 1} = 1140
    \]

The final answer is \( \boxed{1140} \).