Final answer:
The program of four courses from a menu that includes English and at least one mathematics course can be chosen in 9 different ways, as students must choose among combinations of the remaining courses after English and mathematics are accounted for.
Step-by-step explanation:
The student must choose a program of four courses from the given options: English, Algebra, Geometry, History, Art, and Latin. The program must contain English and at least one mathematics course (Algebra or Geometry). Since English is mandatory, we have one course fixed. Now, we need to choose three more courses from the remaining five, keeping in mind that at least one must be a mathematics course.
First, let's choose one mathematics course (either Algebra or Geometry):
we have 2 choices here. After choosing one math course, we have two more courses to choose from the remaining four non-math courses.
We can find the number of combinations using the combination formula C(n, k) = n! / (k!(n-k)!), where 'n' is the total number of options remaining, and 'k' is the number of options to choose.
Using this, we get C(4, 2) = 4! / (2!2!) = 6 ways to choose the other two courses. However, this only accounts for programs with exactly one mathematics course.
If we want to include the possibility of choosing both Algebra and Geometry, we need to add the combinations where both math courses are chosen.
Thus, we would add the number of ways to choose the remaining two courses from the non-math options, which is C(3, 2) = 3.
Finally, the total number of ways to choose the program is the sum of the combinations with one math course and the combinations with two math courses: 6 + 3 = 9.