This is a continuation of Schedule four different exams in a $31$-day period with restriction.
If someone can check my attempts below:
A professor has to schedule four different exams in a $31$-day period. No more than one exam can be scheduled in any one day.
(a) In how many ways can he schedule the exams?
Step $1$: Choose the day for exam $1$: $31$ choices
Step $2$: Choose the day for exam $2$: $30$ choices
Step $3$&$4$: Analogous to previous: $29\cdot 28$ choices.
So there are $31\cdot 30 \cdot 29\cdot 28$ ways to schedule the exams.
However, there are $4!$ ways to arrange the exams themselves. So there are $4!\times (31\cdot 30 \cdot 29\cdot 28)$ ways to arrange the exams.
(b) In how many ways can he schedule the exams if there should be at least three days off between any two finals?
Consider the complement where there are either $0,1$ or $2$ days between any two finals.
Case $1$: $0$ days between any two finals
This would mean the tests are on consecutive days. Consider $E_1 E_2 E_3 E_4$ as one unit. There are $8$ choices to place the exams.
(I drew $31$ spaces for the exams and put slashes after every $4$ spaces.)
Case $2$: $1$ days between any two finals
There are $5$ choices to place the exams.
(Using the $31$ spaces for the exams, I put slashes after every $6$ spaces)
Case $3$: $2$ days between any two finals
Using same approach in previous cases, there are $3$ choices of placing the exams.
Considering there are $4!$ ways to arrange the exams themselves. There are $4!\times (8+5+3)=384$ ways to schedule the exams. (BAD CASES)
So, there are $4!\times (31\cdot 30\cdot 29\cdot 28) - 384$ way to schedule the exams such that there are at least three days between any two finals.
Note: I am not sure this attempt is correct. This is how I understood it.
For (b): There are $4$ black days $\bullet$, then $9$ compulsory white days $\circ$, and $18$ optional white days $\circ$. The black days and the optional white days can be arranged in ${18+4\choose 4}$ ways. For each such arrangement insert three compulsory white days immediately after each of the three first black days in order to obtain an admissible sequence. Finally attribute a particular exam to each of the black days. It follows that there are $${22\choose4}\cdot 4!=175\,560$$ admissible schedules.