A simple permutation question - discrete math There are five distinct computer science books, three distinct mathematics books, and two distinct art books. In how many ways can these books be arranged on a shelf if one of the art books is on the left of all computer science book, and the other art book is on the right of all the computer science books?




For my answer I was thinking since there are $10$ books total and $5$ are computer science books, then I could place an art book on the first, second, third, fourth slot ... but I do not know how to finish it, I am sure it has something to do with rule of sum. Any advice appreciated. Thank you.
You are going in the right direction, but don't think in terms of slots. Just organize the books on the floor, then put them on the shelf.

First, order the computer science books. There are $5!$ ways of putting them in. Then put the two art books on both ends; there are $2$ ways of doing it.

Now you have $7$ books, with eight spaces between and to the left and right of them. We just need to put in the three math books. Let's say they are a Calculus I, a Calculus II, and a Calculus III textbooks.

We put in the Calc I book first. There are 8 places where it can go, relative to the other 7 books. Next, put the Calc II book in; there are 9 places where it can go, relative to the other 8 books that have already been placed. Finally, put in the Calc III book.
answer is $2520$

Number of permutations of $n$ things, taken all at a time, in which $p$ are of one type, $q$ of them are of second-type, $r$ of them are of third-type, and rest are all different is given by
$$
\frac{n!}{p!  q! r!}
$$
so
$$
\frac{10!}{5! 3! 2!} = 2520
$$