A student must choose a program of four courses from a list of courses consisting of English, Algebra, Geometry, History, Art, and Latin. This program must contain English and at least one mathematics course. In how many ways can this program be chosen?
The program must contain exactly $3$ of the $5$ classes that are not English. Therefore, there are $\tbinom{5}{3} = 10$ valid programs if we ignore the mathematics requirement.

Since there are $2$ math classes (Algebra and Geometry), $3$ of the $5$ classes besides English are not math. Therefore, there is only one program that satisfies the English requirement, but does not satisfy the mathematics requirement (the program consisting of English, History, Art, and Latin). It follows that the number of programs satisfying both requirements is $10-1=\boxed{9}.$
9
Level 5
Counting & Probability
Problem: A student must choose a program of four courses from a list of courses consisting of English, Algebra, Geometry, History, Art, and Latin. This program must contain English and at least one mathematics course. In how many ways can this program be chosen?
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To determine the number of ways to choose a program of four courses, we start by identifying the constraints: the program must include English and at least one mathematics course. The available mathematics courses are Algebra and Geometry. 

First, we note that English is mandatory, leaving us with three additional courses to select from the remaining options: Algebra, Geometry, History, Art, and Latin. This gives us a total of five courses to choose from after including English.

Next, we need to ensure that at least one mathematics course is included in the selection. We can approach this by using complementary counting. We will first calculate the total number of ways to choose any three courses from the five available courses (including English) and then subtract the cases where no mathematics course is selected.

The total number of ways to choose three courses from the five available courses is given by the combination formula C(n, k), where n is the total number of items to choose from, and k is the number of items to choose. Thus, we calculate:

C(5, 3) = 5! / (3! * (5-3)!) = 10.

Next, we calculate the number of ways to choose three courses that do not include any mathematics courses. The remaining courses, excluding Algebra and Geometry, are History, Art, and Latin. We can only choose from these three courses, which gives us:

C(3, 3) = 1.

Now, we subtract the number of invalid selections (those without a mathematics course) from the total selections:

Total valid selections = Total selections - Invalid selections = 10 - 1 = 9.

Thus, the final answer is 9.<|END_OF_LATENT|> The final answer is $9$.