jmc

Note that common factors (from $3$ to $20,$ inclusive) of the numerator and the denominator cancel. Therefore, the original expression becomes $$\frac{1}{\cancel{3}}\cdot \frac{2}{\cancel{4}}\cdot \frac{\cancel{3}}{\cancel{5}}\cdots \frac{\cancel{18}}{\cancel{20}}\cdot \frac{\cancel{19}}{21}\cdot \frac{\cancel{20}}{22}=\frac{1\cdot 2}{21\cdot 22}=\frac{1}{21\cdot 11}=\boxed{\frac{1}{231}}.$$