Assume that there are arbitrarily large number of copies of ten different discrete math books. Each students must choose one book. Assume that $500$ students taking the course are indistinguishable.
A) How many different ways can the students choose books, if one specific book must chosen by at least $20$ students.
For this problem, I did $10^{500} - 10^{20}$. Not sure how correct this is.
B) How many different ways can the students choose books, if one specific book must be chosen by at most $20$ students.
Now for this one, the limit is $20$, do I sum the the different ways of the book been chosen by $1$ student up to $20$?
For part A, it's not quite like this.
Let's first assign $20$ students to that specific book. Then we have $480$ students who are free to choose whatever they want. Then, the total number of ways can be calculated using stars and bars to give $\displaystyle \binom{489}{9}$ ways.
For part B, consider the first case where $0$ students choose the specific book. Then, there are $\displaystyle \binom{508}{8}$ ways to split the students among the remaining $9$ books. If there is $1$ student that chooses the specific book, then there are $\displaystyle \binom{507}{8}$ ways to split the remaining students, all the way to $\displaystyle \binom{488}{8}$ .
Can you find a way to sum these?