Let the position give the course to which the professor is assigned question is asking for the arrangments of
$P_1,P_1,P_2,P_2,P_3,P_3,P_4,P_4,P_5,P_5,P_6,P_6,P_7,P_7,P_8,P_8,P_9,P_9,P_{10},P_{10}$
which is
\[
\frac{20!}{2!\times2!\times \ldots \times 2!} = \frac{20!}{(2!)^{10}}
\]
We can also consider as DODB template from 20 potitions select 2 for $P_1$,select 2 for $P_2$,….,select 2 for $P_{10}$
\[
\binom{20}{2} \times \binom{18}{2} \times \ldots \times \binom{2}{2} = \frac{20!}{2!(20-2)!} \times \frac{18!}{2!(18-2)!} \times \ldots \times \frac{2!}{2!(2-2)!}
\]
Simplifying each term:
\[
= \frac{20!}{2! \cdot 18!} \times \frac{18!}{2! \cdot 16!} \times \ldots \times \frac{2!}{2! \cdot 0!}
\]
Now canceling out terms:
\[
= \frac{20 \cdot 19}{2} \times \frac{18 \cdot 17}{2} \times \ldots \times \frac{2.1}{2}= \frac{20!}{(2)^{10}}
\]
