2017 AMC 10B Problem 13

The problem tells us how many students take "at least two" classes, not how many are in each overlap. So sort the students by how many classes each one takes.

1. Let $a$ = students taking exactly one class, $b$ = exactly two, $c$ = all three.
2. Every student takes at least one class, so $a + b + c = 20$.
3. "At least two" means two or three: $b + c = 9$. So $a = 20 - 9 = 11$.
4. Now add up the class sizes: $10 + 13 + 9 = 32$. This counts sign-ups. A student in one class is counted once, a student in two classes twice, a student in three classes three times:
$$a + 2b + 3c = 32.$$
5. Put in $a = 11$: $2b + 3c = 21$.
6. From step 3, $2b + 2c = 18$. Subtract this from $2b + 3c = 21$: $c = 3$.
7. Check: $c = 3$, $b = 6$, $a = 11$. Then $a + b + c = 20$ and $11 + 12 + 9 = 32$.

The answer is $\boxed{\textbf{(C)}\ 3}$.

Adding the class sizes counts each student once for every class they take. When the problem gives "exactly one / two or more" information rather than the pairwise overlaps, counting sign-ups this way gives a small system of equations with no Venn diagram formula needed. Always check which kind of count your given numbers really describe.

Sort students by how many classes they take: let $a$ take exactly one, $b$ exactly two, and $c$ all three.

Every student takes at least one class, so $a + b + c = 20$. Taking at least two classes means $b + c = 9$, hence $a = 11$.

Now count enrollments. Adding the three rosters, $10 + 13 + 9 = 32$, counts each student once per class taken:
$$a + 2b + 3c = 32 \quad\Longrightarrow\quad 11 + 2b + 3c = 32 \quad\Longrightarrow\quad 2b + 3c = 21.$$
Subtract twice $b + c = 9$: $c = 21 - 18 = 3$.

The answer is $\boxed{\textbf{(C)}\ 3}$.