Since you have to choose 20 persons out of a total of 52 persons and there are only 10 professors, so any combination of 20 people from the lot will contain at least 10 associate professors. Therefore what you have to do is find those combinations of 20 persons which have at least 2 professors. Total number of ways of selecting a group of 20 persons is $52 \choose 20$
Now we find the number of groups which have less than 2 professors, i.e., have either 0 or 1 professor in the group of 20 persons. Total number of such combinations are ${42 \choose 20}{10 \choose 0} + {42 \choose 19}{10 \choose 1}$
Therefore, the required number of ways is ${52 \choose 20} - {42 \choose 20} - 10{42 \choose 19} $