A student has to take eight hours of classes a week. He wants to have fewer hours on Friday than on Thursday. In how many ways can he do this?

My solution:

The problem requires the assumption that the number of hours he takes per day must be an integer. Also, assume that he can have days with no classes.

Number of weak compositions of 8 into 5 parts = $\displaystyle \dbinom{12}{4}$ = 495

Let us now find the number of compositions in which he takes an equal number of hours on Thursday and Friday.

Case 1:He does not work on Thursday and Friday. Number of such compositions = Number of weak compositions of 8 into 3 parts = $\displaystyle \dbinom{10}{2}$ = 45

Case 2:He works 1 hour a day on Thursday and Friday. Number of such compositions = Number of weak compositions of 6 into 3 parts = $\displaystyle \dbinom{8}{2}$ = 28

Case 3:He works 2 hours a day on Thursday and Friday. Number of such compositions = Number of weak compositions of 4 into 3 parts = $\displaystyle \dbinom{6}{2}$ = 15

Case 4:He works 3 hours a day on Thursday and Friday. Number of such compositions = Number of weak compositions of 2 into 3 parts = $\displaystyle \dbinom{4}{2}$ = 6

Case 5:He works 4 hours a day on Thurday and Friday. Number of such compositions = 1

So, there are 45 + 28 + 15 + 6 + 1 = 95 compositions in which he takes an equal number of hours on Thursday and Friday.

So, the number of compositions in which he takes anunequal number of hours on Thursday and Friday = 495 - 95 = 400

By symmetry, half of these compositions will have fewer hours on Friday than on Thursday. So, the answer is 200.

Am I right?