inclusion-exclusion: class distribution so that professor teaches the same two courses both semesters 
How many ways are there to assign each of five professors in a math department to two courses in the fall semester (i.e., $10$ different math courses in all) and then assign each professor two courses in the spring semester such that no professor teaches the same two courses both semesters?

My answer is:
$ P(C(10,2),5) - C(5,1) \cdot 10 \cdot P(C(9,2),4) + C(5,2) \cdot (10 \cdot 9) \cdot P(C(8,2),3) - ... $
And I am wrong. Why am I wrong? You have 10 courses, and 5 professors are supposed to pick one combination of the courses such that (fall, spring) - so it's $P(C(10,2),5)$. By excluding each case of one professor having the same courses for two semesters in a row, it becomes the following sections of the equation $ - C(5,1) \cdot 10 \cdot P(C(9,2),4) + C(5,2) \cdot (10 \cdot 9) \cdot P(C(8,2),3)$ and so on. 
What's wrong with this logic?

The number of ways five professors can each be assigned two courses in the fall semester is 
$$\binom{10}{2}\binom{8}{2}\binom{6}{2}\binom{4}{2}\binom{2}{2}$$
To see this, list the professors in some order, say by seniority.  Assign the first professor on the list two of the ten courses, which leaves eight courses available.  Assign the next professor on the list two of the remaining eight courses and so forth.  
We need to multiply this by the number of ways of assigning each professor two courses in the spring semester without assigning any of those professors the same two courses that the professor taught in the fall semester.
A professor teaches the same two courses both semesters:  There are five ways to choose which professor is assigned the same two courses that he or she taught in the fall semester.  The remaining eight courses are distributed among the remaining four professors in 
$$\binom{8}{2}\binom{6}{2}\binom{4}{2}\binom{2}{2}$$
ways.  Thus, there are 
$$\binom{5}{1}\binom{8}{2}\binom{6}{2}\binom{4}{2}\binom{2}{2}$$
such spring semester assignments. 
There are 
$$\binom{5}{2}\binom{6}{2}\binom{4}{2}\binom{2}{2}$$ 
ways to assign two professors to teach the same two courses they taught in the fall semester.
There are 
$$\binom{5}{3}\binom{4}{2}\binom{2}{2}$$
ways to assign three professors to teach the same two courses they taught in the fall semester.
There are 
$$\binom{5}{4}\binom{2}{2}$$
ways to assign four professors to teach the same two courses they taught in the fall semester.
There are 
$$\binom{5}{5}$$
ways to assign all five professors to teach the same two courses they taught in the fall semester. 
By the Inclusion-Exclusion Principle, the number of admissible spring semester teaching assignments is 
$$\binom{10}{2}\binom{8}{2}\binom{6}{2}\binom{4}{2}\binom{2}{2} - \binom{5}{1}\binom{8}{2}\binom{6}{2}\binom{4}{2}\binom{2}{2} + \binom{5}{2}\binom{6}{2}\binom{4}{2}\binom{2}{2} - \binom{5}{3}\binom{4}{2}\binom{2}{2} + \binom{5}{4}\binom{2}{2} - \binom{5}{5}$$
Therefore, the number of ways five professors can each be assigned two courses in the fall semester and two courses in the spring professor so that no professor teaches the same two courses both semesters is 
$$\binom{10}{2}\binom{8}{2}\binom{6}{2}\binom{4}{2}\binom{2}{2}\left[\binom{10}{2}\binom{8}{2}\binom{6}{2}\binom{4}{2}\binom{2}{2} - \binom{5}{1}\binom{8}{2}\binom{6}{2}\binom{4}{2}\binom{2}{2} + \binom{5}{2}\binom{6}{2}\binom{4}{2}\binom{2}{2} - \binom{5}{3}\binom{4}{2}\binom{2}{2} + \binom{5}{4}\binom{2}{2} - \binom{5}{5}\right]$$
