For how many ordered pairs of positive integers $(a, b)$ is $1<a+b<22$ ?
If $a=1$, we get $1<1+b<22$ or $0<b<21$, which means that $b$ can equal $1,2,3, \ldots, 19,20$.

If $a=2$, we get $1<2+b<22$ or $-1<b<20$.

Since $b$ is positive, then $b$ can equal $1,2,3, \ldots, 18,19$.

In general, $1-a<b<22-a$. Since $b$ is positive, then $b$ satisfies $1 \leq b \leq 21-a$, which means that there are $21-a$ possible values for $b$ for a given $a$.

As $a$ runs from 1 to 20 , there are thus $20,19,18, \ldots, 3,2,1$ possible values for $b$ in these 20 cases. This means that the total number of pairs $(a, b)$ is $20+19+18+\cdots+3+2+1=\frac{1}{2}(20)(21)=210$.