AMC 10 · 2013 · #7

Grade 7 counting
Archetype Overlap / Complement Counting └ Sub-archetype Complement by Subtracting Forbidden Arrangements
combinations-basiccomplementary-countingcasework complementary-countingcasework ↑ Prerequisites: combinations-basic
📏 Medium solution 💡 2 insights
Problem
Pick 4 courses from a list of 6: English, Algebra, Geometry, History, Art, and Latin. The four chosen courses must include English and must include at least one mathematics course (Algebra or Geometry). Count how many different four-course programs are possible.

Pick an answer.

(A)
6
(B)
8
(C)
9
(D)
12
(E)
16

AMC 10 2013 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Change Focus / Count the Complement

The phrase "at least one math course" is easier to handle through its opposite. First place the forced English course, then count every way to fill the other three seats, and finally subtract the few programs that have no math course at all.

1STEP 1

Lock English, three seats left

English is forced, so place it first; three seats remain, to be filled from the five other courses.

Fill 3 seats from {Algebra, Geometry, History, Art, Latin}
2STEP 2

Count every three-course fill

Ignore the math rule for a moment: choosing 3 of the 5 leftover courses gives 10 groups.

C(5, 3)=(5 · 4 · 3)/(3 · 2 · 1)=10
3STEP 3

Count the no-math programs

The rule fails only with zero math, forcing History, Art, Latin — just 1 bad group.

C(3, 3)=1 (History, Art, Latin)
4STEP 4

Subtract the forbidden program

Remove that one bad group from the 10, leaving 9 valid programs, so the answer is (C).

10-1=9
Answer
9
Direct casework agrees: with exactly one math course, choose Algebra or Geometry (2 ways) then 2 of {History, Art, Latin} (3 ways) = 6; with both math courses, choose 1 of the 3 non-math courses = 3. Total 6 + 3 = 9, matching the complement count and confirming (C).
💡Key takeaway

When a rule says "at least one," count everything, then subtract the cases that have none.

  • Lock English, three seats left
  • Count every three-course fill
  • Count the no-math programs
  • Subtract the forbidden program

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