AMC 10 · 2025 · #14

Grade 7 probability
Archetype Branching State Enumeration └ Sub-archetype Constrained Sequence or Path Count
probability-basiccombinations-basicpermutations-basic identify-subproblems ↑ Prerequisites: probability-basic
📏 Medium solution 💡 3 insights
Problem
Nine athletes of all different heights each draw a colored wristband at random from a bag holding 3 blue, 3 red, and 3 green bands, splitting them into three groups of three. The tallest person in each group is that group's captain. Find the probability that the three group captains turn out to be the three tallest athletes overall.

Pick an answer.

(A)
$\dfrac{2}{9}$
(B)
$\dfrac{2}{7}$
(C)
$\dfrac{9}{28}$
(D)
$\dfrac{1}{3}$
(E)
$\dfrac{3}{8}$

AMC 10 2025 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Change Focus / Count the Complement

Chasing captains directly is messy. Tool #16 (Change Focus) rewrites the event into something equivalent but simple: the three tallest are captains exactly when no two of them share a group. Once the event is about only the top three athletes, Tool #9 (Solve an Easier Related Problem) throws away the other six people entirely — they never affect who captains whom. Tool #7 (Identify Subproblems) then handles the top three one draw at a time: the tallest lands anywhere, then ask whether the second and third each grab a fresh color.

1STEP 1

Rewrite the winning condition

A top-three athlete loses only to another top-three athlete, so all three captain exactly when they land in three different groups.

captains = {top 3} ⇔ top 3 all in different groups
2STEP 2

Track only the three tallest

The other six change nothing, so ignore them and let the three tallest draw first — the tallest is a captain whichever color he pulls.

P(tallest gets a captaincy) = 1
3STEP 3

Second tallest needs a new color

After the tallest draws, 8 bands remain and only 2 share his color, so 6 of the 8 keep the second-tallest apart from him.

P(second differs) = 6/8
4STEP 4

Third tallest needs the last color

Two colors are taken, so the third-tallest must draw the untouched one: 3 of the 7 bands left do that.

P(third differs) = 3/7
5STEP 5

Multiply the three chances

The three groupings happen in sequence, so multiply the probabilities of each success and simplify the fraction.

1 × 6/8 × 3/7 = 18/56 = 9/28 → (C)
Answer
9/28
Cross-check by whole-line counting: lay all nine athletes in a row where the first three positions are one color, the next three another, the last three the third. There are 9! orderings. For a favorable one, place the three tallest in three different color-blocks: 3! ways to pick which block each takes, 3×3×3 ways to choose their exact seat inside their block, and 6! ways to seat the rest. That gives (3! · 27 · 6!)/9! = (6 · 27)/(9 · 8 · 7) = 162/504 = 9/28, matching (C). The value is a little under 1/3, which is sensible: it is fairly but not overwhelmingly likely the three tallest scatter.
💡Key takeaway

The three tallest are captains only when they land in three different colors, so just track those three and multiply the chances of each grabbing a fresh color.

  • Rewrite the winning condition
  • Track only the three tallest
  • Second tallest needs a new color
  • Third tallest needs the last color
  • Multiply the three chances

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