A three-person committee must be chosen from a group of 7 professors

@anujkch,
The answer to your question first:

When you do 7C1*16C2, you make a critical error. Consider the 7 professors to be A,B,C,D,E,F,G. Supposing you select A to be separate (i.e. part of the 7C1 selection) and the other two professors to be selected from B through G. You will then get a case A(BC). Now consider the case where you select B as part of the separate case, and choose the other two professors from A,C,D,E,F,G,H. Again you will get a case B(AC), which is the same as ABC in terms of selection. Therefore your method causes double counting at places and gives you an answer which is greater than the actual one.

@ankitbansal85,
The error you make is different. In the PPP case, 7*6*5 is the number of ways you can arrange, and not just select three professors from 7 professors (i.e. 7*6*5 = 7P3). The number of ways to select 3 professors from 7 professors is 7C3. Similarly, in the PSS case, you must use 7*10C2, not 7*10*9, the latter is the number of ways to arrange 1 professor and 2 students in a row, not just select them. Correct this approach and you will get the right answer.

As for the right answer, it can be obtained in multiple ways,
Method 1:
7C1 * 10C2 + 7C2 * 10C1 + 7C3 = 315 + 210 + 35 = 560

Method 2:
Total number of ways to select at least one professor = Number of ways to select 3 people from 10 students and 7 professors - Number of ways to select 3 people from 10 student (i.e. without including any professor)
= 17C3 - 10C3 = 680 - 120 = 560