imc

Let us split this up into two cases. Case $1$: The student chooses both algebra and geometry. This means that $3$ courses have already been chosen. We have $3$ more options for the last course, so there are $3$ possibilities here. Case $2$: The student chooses one or the other. Here, we simply count how many ways we can do one, multiply by $2$, and then add to the previous. Assume the mathematics course is algebra. This means that we can choose $2$ of History, Art, and Latin, which is simply $\left(\genfrac{}{}{0ex}{}{3}{2}\right)=3$. If it is geometry, we have another $3$ options, so we have a total of $6$ options if only one mathematics course is chosen. Thus, overall, we can choose a program in $6+3=\boxed{9}$ ways