2025 AMC 12B Problem 11

Question

Nine athletes, no two of whom are the same height, try out for the basketball team. One at a time, they draw a wristband at random, without replacement, from a bag containing

3

blue bands,

3

red bands, and

3

green bands. They are divided into a blue group, a red group, and a green group. The tallest member of each group is named the group captain. What is the probability that the group captains are the three tallest athletes?

\dfrac{2}{9}

\dfrac{2}{7}

\dfrac{9}{28}

\dfrac{1}{3}

\dfrac{3}{8}

Answer: C

Concepts:basic probability symmetry

Difficulty rating: 1590

Small hint:

The captains are the three tallest exactly when the three tallest land in three different groups

Big hint:

Place the tallest three one at a time; each must avoid the groups already used, whose slots shrink the pool

Solution:

Each group has

3

slots. The three tallest athletes are the captains precisely when they fall into three different groups. Placing them one at a time into the

9

slots, the second must avoid the first’s group (

6

of the remaining

8

slots) and the third must avoid both used groups (

3