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2013 AMC 10A Problem 11

2013 AMC 10A Problem 11AMC 10 Step by Step

Name: Date: September 13, 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

C(n,2) = 10 forces n = 5 council members, and C(5,3) = C(5,2) = 10 by the symmetry of binomial coefficients.

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 $\boxed{\textbf{(A)}\ 10}$ .

Why this works

The problem hides the group size behind a count; recovering $n$ from $\binom{n}{2}$ is the only real step. After that, $\binom{5}{3} = \binom{5}{2}$ because choosing the $3$ people who serve is the same as choosing the $2$ who do not, which is why the two counts coincide.

The trap

Assuming the two committees must be disjoint and computing C(3,3) or similar, when the problem says members may serve on both.

Common mistakes

Assuming the two committees must be disjoint and computing C(3,3) or similar, when the problem says members may serve on both.
Using $n = 10$ as the council size and answering $\binom{10}{3} = 120$ (not even a choice, but it wastes time).

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