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2017 AMC 10B Problem 13

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2017 AMC 10B Problem 13AMC 10 Step by Step
Name:                             Date: September 13, 2026
Inclusion-Exclusion · difficulty 3 of 5 · about 4 min · position P11-15

There are 2020 students participating in an after-school program offering classes in yoga, bridge, and painting. Each student must take at least one of these three classes, but may take two or all three. There are 1010 students taking yoga, 1313 taking bridge, and 99 taking painting. There are 99 students taking at least two classes. How many students are taking all three classes?

  1. A)

    11

  2. B)

    22

  3. C)

    33

  4. D)

    44

  5. E)

    55

Answer

C

Key insight

Count by number of classes taken: a + b + c = 20, b + c = 9, and the roster total gives a + 2b + 3c = 32.

Solution

Sort students by how many classes they take: let aa take exactly one, bb exactly two, and cc all three.

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

Now count enrollments. Adding the three rosters, 10+13+9=3210 + 13 + 9 = 32, counts each student once per class taken:
a+2b+3c=32⟹11+2b+3c=32⟹2b+3c=21.a + 2b + 3c = 32 \quad\Longrightarrow\quad 11 + 2b + 3c = 32 \quad\Longrightarrow\quad 2b + 3c = 21.
Subtract twice b+c=9b + c = 9: c=21−18=3c = 21 - 18 = 3.

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

Why this works

When the data are "how many take at least two" rather than the individual pairwise overlaps, the standard inclusion-exclusion formula is awkward. Counting by multiplicity (exactly one, two, or three classes) turns the roster total into a weighted sum and gives a tiny linear system. Look for the version of the count that matches the information you are actually given.

The trap

Applying |A union B union C| = sum of singles minus pairs plus triple with the wrong meaning of the 9, which counts students in at least two classes, not pairwise overlaps.

Common mistakes

  • Applying |A union B union C| = sum of singles minus pairs plus triple with the wrong meaning of the 9, which counts students in at least two classes, not pairwise overlaps.
  • Forgetting that students in all three classes are counted three times in the roster total, and writing a+2b+2c=32a + 2b + 2c = 32.

Techniques

Set up the equation/formula and compute; no special trick needed