How many ways can you arrange 3 French books, 2 history books, and 5 math books if (a) all books are different? (b) all French books look the same, all history books look the same, and all math books look the same?

The number of ways is $N!$.

$$\text{Number of ways} = 10!$$

$$10! = 10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 3,628,800$$

The number of ways to arrange the books if all books are different is $\boxed{3,628,800}$.

$$N = 10$$

Number of French books ($n_F$) = 3
Number of History books ($n_H$) = 2
Number of Math books ($n_M$) = 5

$$\text{Number of ways} = \frac{10!}{3! 2! 5!}$$ $$\text{Number of ways} = \frac{10 \times 9 \times 8 \times 7 \times 6 \times 5!}{ (3 \times 2 \times 1) \times (2 \times 1) \times 5!}$$ $$\text{Number of ways} = \frac{10 \times 9 \times 8 \times 7 \times 6}{3 \times 2 \times 1 \times 2 \times 1}$$ $$\text{Number of ways} = \frac{30240}{12}$$ $$\text{Number of ways} = 2520$$

The number of ways to arrange the books with identical books in each category is $\boxed{2520}$.