2013 AMC 10A Problem 7

1. English is required. Put it in first. Now the student picks $3$ more courses from the other $5$: Algebra, Geometry, History, Art, Latin.
2. Count all ways to pick 3 of 5. Picking the $3$ to take is the same as picking the $2$ to leave out, and there are $10$ pairs you could leave out. The formula gives the same number:
$$\binom{5}{3} = \frac{5 \times 4 \times 3}{3 \times 2 \times 1} = 10.$$
3. Throw out the bad ones. A program is bad if it has no math course. That means all $3$ extra courses are History, Art and Latin. There is only $1$ such program.
4. Answer. $10 - 1 = 9$ programs.

Check by listing. Programs with exactly one math course: choose Algebra or Geometry ($2$ ways), then $2$ of History, Art, Latin ($3$ ways), for $6$. Programs with both math courses: add $1$ of History, Art, Latin, for $3$. Total $6 + 3 = 9$.

The answer is $\boxed{\textbf{(C)}\ 9}$.

"At least one" is usually easier the other way around: count everything, then subtract the choices that have none. Putting the required course in first makes the problem smaller. The listing check in the solution splits the programs into "one math course" and "two math courses", so no program gets counted twice.

English is required, so the student really chooses $3$ more courses from the other $5$: Algebra, Geometry, History, Art, Latin. That can be done in $\binom{5}{3} = 10$ ways.

The only forbidden selections are those with no mathematics course, i.e. all three courses from $\{$History, Art, Latin$\}$. There is exactly $\binom{3}{3} = 1$ such selection.

Valid programs: $10 - 1 = 9$. The answer is $\boxed{\textbf{(C)}\ 9}$.