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2013 AMC 10A Problem 7

2013 AMC 10A Problem 7AMC 10 Step by Step

Name: Date: September 13, 2026

Basic Counting · difficulty 2 of 5 · about 2 min · position P1-10

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?

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

Answer

Key insight

English is forced, so choose 3 of the other 5 courses and throw away the single choice with no math course.

Solution

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 $\boxed{\textbf{(C)}\ 9}$ .

Why this works

"At least one" conditions are usually easiest through the complement: count everything, subtract the selections with none. Fixing the mandatory item first shrinks the problem to a plain combination count. The direct route also works here (one math course: $2 \cdot \binom{3}{2} = 6$ ; both math courses: $\binom{3}{1} = 3$ ; total $9$ ), but the complement is faster and less error-prone.

The trap

Counting 'at least one math course' by picking a math course first and then any two others, which double-counts programs with both math courses.

Common mistakes

Counting 'at least one math course' by picking a math course first and then any two others, which double-counts programs with both math courses.
Forgetting that English is already one of the four courses and choosing $4$ from the remaining $5$ .

Techniques

Count the complement and subtract from the total