AMC 10 · 2021 · #22

Grade 7 probability

Archetype Overlap / Complement Counting └ Sub-archetype At-Least-One Complement: 1 − P(None)

principle-of-inclusion-exclusionprobability-basicpermutations-basiccombinations-basic complementary-countingidentify-subproblemssystematic-enumeration ↑ Prerequisites: probability-basiccombinations-basic

📏 Long solution 💡 4 insights

Problem

Ang, Ben, and Jasmin each have 5 distinctly-colored blocks (red, blue, yellow, white, green). Each person independently and uniformly at random places one block in each of 5 empty boxes (so each person uses a random permutation). Find the probability that at least one box ends up holding three blocks of the same color (one from each person). The answer is

\frac{m}{n}

in lowest terms; report m + n.

Pick an answer.

(A)

~47

(B)

~94

(C)

~227

(D)

~471

(E)

~542

AMC 10 2021 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Change Focus / Count the Complement

Tool #16 (Change Focus) — "at least one" is the classic Inclusion-Exclusion trigger; we'll count by inclusion-exclusion over the 5 events A_i = "box i is uniform-colored". Tool #9 (Easier Problem) — fix Ang's placement WLOG (this is symmetry, not loss of generality) so we only need to count valid (Ben, Jasmin) pairs out of (5!)². Tool #7 (Subproblems) — compute each |A_i₁ ∩ … ∩ A_i_k| separately as a clean factorial expression. Tool #2 (Systematic List) — write out the five PIE terms and add with signs. Tool #3 (Eliminate) — match m + n against the five answer choices.

1STEP 1

The sample space is (5!)³, but by symmetry we may fix Ang's permutation, leaving just 14,400 ordered (Ben, Jasmin) pairs to count.

|Ω| = (5!)³, fix Ang → effective |Ω'| = (5!)² = 14400

2STEP 2

A box is uniform only when Ben and Jasmin both copy Ang there; this overlapping 'at least one' union is what Inclusion-Exclusion counts.

|A₁ ∪ … ∪ A₅| = S₁ - S₂ + S₃ - S₄ + S₅

3STEP 3

Forcing k boxes to match costs Ben and Jasmin (5-k)! each, so the intersection is ((5-k)!)² and S_k = C(5, k)((5-k)!)².

|A_i₁ ∩ … ∩ A_i_k| = ((5 - k)!)²; S_k = C(5, k)((5-k)!)²

4STEP 4

Plug in k = 1…5: S₁ = 2880, S₂ = 360, S₃ = 40, S₄ = 5, S₅ = 1.

S₁ = 2880, S₂ = 360, S₃ = 40, S₄ = 5, S₅ = 1

5STEP 5

PIE gives N = 2880 - 360 + 40 - 5 + 1 = 2556, so P = $\frac{2556}{14400}$ .

N = 2880 - 360 + 40 - 5 + 1 = 2556; P =

\frac{2556}{14400}

6STEP 6

Since 2556 = 2²·3²·71 and 14400 = 2⁶·3²·5², dividing by gcd 36 reaches $\frac{71}{400}$ in lowest terms, so m = 71, n = 400.

\frac{2556}{14400}

\frac{71}{400}

, m = 71, n = 400

7STEP 7

Add: m + n = 71 + 400 = 471, matching choice (D).

m + n = 71 + 400 = 471 → (D)

Answer

~471

Sanity. Probability

\frac{71}{400}

≈ 0.1775, so about 18% of arrangements have at least one same-color box. That feels right: for a single box, P(A_i) =

\frac{1}{5}

\frac{1}{5}

\frac{1}{25}

= 0.04. With 5 boxes, a naive overestimate (ignoring overlap) is 5 ·

\frac{1}{25}

= 0.20, and the true value 0.1775 is just slightly less — consistent with PIE removing the small overcount. Also S₁ - S₂ = 2520, so the first two PIE terms dominate; S₃, S₄, S₅ contribute +36 total — small corrections, as expected when individual events are rare.

💡Key takeaway

This hard AMC 10 problem only needs Grade 7 probability you already know — "at least one" plus inclusion-exclusion. Fix Ang's blocks first (symmetry costs nothing); then for each subset of k boxes, force Ben and Jasmin to copy Ang there: C(5, k)((5-k)!)² ways. Alternate signs to get 2880 - 360 + 40 - 5 + 1 = 2556 favorable out of (5!)² = 14400, reduce to $\frac{71}{400}$ , and m + n = 471.

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