In how many ways can you choose 3 books from a shelf containing 7 math books, 5 physics books, and 4 chemistry books if at least one book must be a math book?
First, let's calculate the total number of ways to choose any 3 books without restrictions. There are 16 books in total (7 math + 5 physics + 4 chemistry). We can use the combination formula:

C(n, k) = n! / (k!(n-k)!)

where n is the total number of items and k is the number of items to choose.

Total combinations = C(16, 3) = 16! / (3!(16-3)!)
                             = 560

Now let's calculate the number of ways to choose 3 books without any math book. This means we are choosing from 9 books (5 physics + 4 chemistry).

Combinations without a math book = C(9, 3) = 9! / (3!(9-3)!)
                                                = 84

To find the number of ways to choose 3 books with at least one math book, we subtract the combinations without a math book from the total combinations:

Combinations with at least one math book = Total combinations - Combinations without a math book
                                                        = 560 - 84
                                                        = 476

There are 476 different ways to choose 3 books with at least one math book.