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

A student must choose a program of four courses from a menu of courses consisting of English, Algebra, Geometry, History, Art, and Latin. This program must contain English and at least one mathematics course. In how many ways can this program be chosen?

Pick an answer.

(A)

(B)

(C)

(D)

(E)

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

Try it yourself first — the explanation is most useful after you’ve attempted it.

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Toolkit + CCSS Solution

Understand

Restated: 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.

Givens: Six courses are available: English, Algebra, Geometry, History, Art, Latin.; Exactly four courses are chosen for the program.; Algebra and Geometry are the two mathematics courses.

Unknowns: The number of valid four-course programs.

Understand

Givens: Six courses are available: English, Algebra, Geometry, History, Art, Latin.; Exactly four courses are chosen for the program.; Algebra and Geometry are the two mathematics courses.

Plan

Primary tool: #16 Change Focus / Count the Complement

Secondary: #2 Make a Systematic List, #7 Identify Subproblems

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.

Execute — Answer: C

#16 Change Focus / Count the Complement 7.SP.C.8 Step 1

Lock English, three seats left

English must be in every program, so place it first.
That fills one of the four seats and removes English from the pool.
The remaining job is to fill three seats using the five leftover courses: Algebra, Geometry, History, Art, and Latin.

$$\text{Fill } 3 \text{ seats from } \{\text{Algebra, Geometry, History, Art, Latin}\}$$

💡 A forced item is free to place, so set it aside and solve the smaller problem.

#2 Make a Systematic List 7.SP.C.8 Step 2

Count every three-course fill

Ignore the math rule for a moment and count every way to choose 3 of the 5 leftover courses.
Listing the groups by hand, or using the choose formula, gives 10 possibilities.

$$\binom{5}{3}=\dfrac{5\cdot 4\cdot 3}{3\cdot 2\cdot 1}=10$$

💡 Counting all groups first is simpler than counting only the allowed ones.

#16 Change Focus / Count the Complement 7.SP.C.8 Step 3

Count the no-math programs

A program breaks the rule only when it has zero math courses.
Then all three remaining seats come from the non-math courses History, Art, and Latin.
There are exactly three such courses, so there is only one way to pick all three of them.

$$\binom{3}{3}=1 \quad (\text{History, Art, Latin})$$

💡 Only one group of three can avoid both math courses, because just three non-math courses exist.

#16 Change Focus / Count the Complement 1.OA.C.6 Step 4

Subtract the forbidden program

Take all 10 programs and remove the single one that has no math course.
That leaves 9 programs that include English and at least one math course, so the answer is (C).

$$10-1=9$$

💡 Everything minus the not-allowed leaves exactly the allowed.

[1] #16 7.SP.C.8 English must be in every program, so place it first. That fills one of the four

[2] #2 7.SP.C.8 Ignore the math rule for a moment and count every way to choose 3 of the 5 lefto

[3] #16 7.SP.C.8 A program breaks the rule only when it has zero math courses. Then all three rem

[4] #16 1.OA.C.6 Take all 10 programs and remove the single one that has no math course. That lea

Review

Reasonableness: 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).

Alternative: Split into cases by how many math courses are included (exactly one, or both) and add the case counts instead of subtracting the complement.

CCSS standards used (min grade 7)

7.SP.C.8 Find probabilities of compound events using organized lists, tables, and simulation (Organizing and counting all possible four-course programs as a sample space, then isolating the no-math programs)
1.OA.C.6 Add and subtract within 20 using strategies (Subtracting the one forbidden program from the ten total programs)

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

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: C

Lock English, three seats left

Count every three-course fill

Count the no-math programs

Subtract the forbidden program

Review

CCSS standards used (min grade 7)

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