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2013 AMC 10A Problem 11

2013 AMC 10A Problem 11AMC 10 Step by Step

Name: Date: October 28, 2026

Basic Counting · difficulty 2 of 5 · about 2 min · position P11-15

A student council must select a two-person welcoming committee and a three-person planning committee from among its members. There are exactly 10 ways to select a two-person team for the welcoming committee. It is possible for students to serve on both committees. In how many different ways can a three-person planning committee be selected?

A) $10$
B) $12$
C) $15$
D) $18$
E) $25$

Answer

Key insight

Ten ways to pick 2 people means the council has 5 members. Then count the ways to pick 3 of those 5, which is also 10.

Solution

The problem hides how many people are on the council. Find that first.

Count pairs for a few sizes. The number of ways to pick $2$ people from a group:

People	3	4	5	6
Pairs	3	6	10	15

(With $n$ people it is $\frac{n(n-1)}{2}$ . For $5$ people: $\frac{5 \times 4}{2} = 10$ .)
2. So the council has 5 members. That is the only size with exactly $10$ pairs.
3. Now pick 3 of the 5. The committees may share people, so the planning committee is any $3$ of the $5$ members.
4. Count. Choosing $3$ people to be on the committee is the same as choosing the $2$ people who are left off. There are $10$ ways to choose those $2$ (Step 1). So there are $10$ ways.

The answer is $\boxed{\textbf{(A)}\ 10}$ .

Why this works

The main step is working backwards from a count to the group size. After that, notice that picking who is in a group of $3$ is the same as picking who is out (a group of $2$ ). That is why both counts are $10$ : $\binom{5}{3} = \binom{5}{2}$ .

The trap

Thinking the two committees must use different people. The problem says a student may be on both, so the 3 can be any 3 of the 5.

Concise solution

Let the council have $n$ members. The number of two-person teams is $\binom{n}{2} = \frac{n(n-1)}{2}$ , and we are told this equals $10$ , so $n(n-1) = 20$ . Since $5 \cdot 4 = 20$ , the council has $n = 5$ members.

Because overlap between the committees is allowed, the planning committee is simply any $3$ of the $5$ members:
$\binom{5}{3} = \frac{5 \cdot 4 \cdot 3}{3 \cdot 2 \cdot 1} = 10.$

The answer is $\boxed{\textbf{(A)}\ 10}$ .

Review in the AoPS books

Introduction to Counting & Probability, Chapter 1: Counting Is Arithmetic
Introduction to Counting & Probability, Chapter 2: Basic Counting Techniques
Introduction to Counting & Probability, Chapter 4: Committees and Combinations

Techniques

Set up the equation/formula and compute; no special trick needed · Test small/specific values or special cases to find or verify the answer