Solution:

Our sequence starts with a $0$ then has sequences of $110$ and $10$ in some order, where they each come after a $0.$

Let the number of $110$ be $x$ and let the number of $10$ be $y.$ Then the number of terms in the sequence is $$3x+2y+1=19,$$ making $$3x+2y=18.$$ The only possible ordered pairs are $$\begin{align*}(x,y) = &(6,0),\\&(4,3),\\&(2,6),\\&(0,9).\end{align*}$$ Then, the number of ways to order the $x+y$ blocks is $$\binom{x+y}x,$$ since we choose which $x$ of the $x+y$ positions hold a $110$ block.

Therefore, the total number of ways is $$\binom 66 + \binom 74 + \binom 82 + \binom 90$$$$=1+35+28+1$$$$=65.$$

Thus, the answer is C .