AMC 10 · 2021 · #22

Grade 7 probability

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 $5$ blocks, colored red, blue, yellow, white, and green; and there are $5$ empty boxes. Each of the people randomly and independently of the other two people places one of their blocks into each box. The probability that at least one box receives $3$ blocks all of the same color is $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. What is $m + n ?$

Pick an answer.

(A)

~47

(B)

~94

(C)

~227

(D)

~471

(E)

~542

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Toolkit + CCSS Solution

Understand

Restated: 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 $\tfrac{m}{n}$ in lowest terms; report $m + n$.

Givens: $3$ people: Ang, Ben, Jasmin — each holds the same $5$ colors of block; $5$ boxes, each receives exactly one block from each person; Each person's placement is a uniformly random permutation of $5$ colors into $5$ boxes; All three placements are independent; Answer choices: (A) $47$, (B) $94$, (C) $227$, (D) $471$, (E) $542$

Unknowns: Probability that at least one box receives $3$ same-color blocks, as $\tfrac{m}{n}$ in lowest terms; $m + n$

Understand

Plan

Primary tool: #16 Change Focus / Count the Complement

Secondary: #2 Make a Systematic List, #9 Solve an Easier Related Problem, #7 Identify Subproblems, #3 Eliminate Possibilities

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 $(\text{Ben}, \text{Jasmin})$ pairs out of $(5!)^2$. Tool #7 (Subproblems) — compute each $|A_{i_1} \cap \cdots \cap A_{i_k}|$ 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.

Execute — Answer: D

#9 Solve an Easier Related Problem 7.SP.C.7 Step 1

Set up the sample space using symmetry.
Each of the three people independently picks a permutation of $5$ colors into $5$ boxes — so the sample space has size $(5!)^3 = 120^3$.
By symmetry we may *fix* Ang's permutation (say Ang puts color $c_i$ in box $i$ for each $i$); this is Tool #9 — we are solving a smaller, equivalent problem.
Now we only need to count favorable $(\text{Ben}, \text{Jasmin})$ pairs out of $(5!)^2 = 14{,}400$ ordered pairs.

$$|\Omega| = (5!)^3,\;\text{fix Ang} \Rightarrow \text{effective } |\Omega'| = (5!)^2 = 14400$$

💡 Fixing Ang doesn't change the probability — it just relabels the colors to match the boxes.

#16 Change Focus / Count the Complement 7.SP.C.8 Step 2

Define event $A_i = $ "box $i$ ends up with three blocks of the same color" (for $i = 1, \ldots, 5$).
Because Ang's color $c_i$ is in box $i$, $A_i$ happens exactly when Ben puts $c_i$ in box $i$ AND Jasmin puts $c_i$ in box $i$.
We want $P(A_1 \cup A_2 \cup \cdots \cup A_5)$.
By the Inclusion-Exclusion Principle (PIE), $|A_1 \cup \cdots \cup A_5| = \sum_k (-1)^{k+1} S_k$ where $S_k = \sum_{|I| = k} |A_{i_1} \cap \cdots \cap A_{i_k}|$.

$$|A_1 \cup \cdots \cup A_5| = S_1 - S_2 + S_3 - S_4 + S_5$$

💡 "At least one" with overlap — exactly when PIE is the right counter.

#7 Identify Subproblems 7.SP.C.8 Step 3

Compute $|A_{i_1} \cap \cdots \cap A_{i_k}|$ for any specific $k$-subset of boxes.
The condition forces Ben to place $c_{i_1}, \ldots, c_{i_k}$ in those $k$ boxes (1 way), and the remaining $5 - k$ colors go in the remaining $5 - k$ boxes in $(5 - k)!$ ways.
Same for Jasmin.
So $|A_{i_1} \cap \cdots \cap A_{i_k}| = ((5 - k)!)^2$ regardless of which $k$-subset.
Therefore $S_k = \binom{5}{k} ((5 - k)!)^2$.

$$|A_{i_1} \cap \cdots \cap A_{i_k}| = ((5 - k)!)^2;\quad S_k = \binom{5}{k}((5-k)!)^2$$

💡 Forcing $k$ matches costs $k$ degrees of freedom for each of Ben and Jasmin — clean factorial squared.

#2 Make a Systematic List 7.SP.C.8 Step 4

List all five terms (Tool #2).
For $k = 1$: $S_1 = \binom{5}{1}(4!)^2 = 5 \cdot 576 = 2880$.
For $k = 2$: $S_2 = \binom{5}{2}(3!)^2 = 10 \cdot 36 = 360$.
For $k = 3$: $S_3 = \binom{5}{3}(2!)^2 = 10 \cdot 4 = 40$.
For $k = 4$: $S_4 = \binom{5}{4}(1!)^2 = 5 \cdot 1 = 5$.
For $k = 5$: $S_5 = \binom{5}{5}(0!)^2 = 1 \cdot 1 = 1$.

$$S_1 = 2880,\; S_2 = 360,\; S_3 = 40,\; S_4 = 5,\; S_5 = 1$$

💡 Each $S_k$ is a single clean product — no overlap to worry about until we add with signs.

#16 Change Focus / Count the Complement 6.NS.B.3 Step 5

Apply PIE: $|A_1 \cup \cdots \cup A_5| = 2880 - 360 + 40 - 5 + 1 = 2556$.
So the probability is $P = \dfrac{2556}{14400}$.

$$N = 2880 - 360 + 40 - 5 + 1 = 2556;\quad P = \dfrac{2556}{14400}$$

💡 Alternating signs cancel the over-counting from each pairwise overlap.

#7 Identify Subproblems 6.NS.B.4 Step 6

Reduce $\dfrac{2556}{14400}$ to lowest terms.
$\gcd(2556, 14400)$: $2556 = 4 \cdot 639 = 4 \cdot 9 \cdot 71 = 2^2 \cdot 3^2 \cdot 71$.
$14400 = 144 \cdot 100 = 2^6 \cdot 3^2 \cdot 5^2$.
$\gcd = 2^2 \cdot 3^2 = 36$.
So $\dfrac{2556}{14400} = \dfrac{2556 / 36}{14400 / 36} = \dfrac{71}{400}$.
Since $71$ is prime and $400 = 2^4 \cdot 5^2$ has no factor of $71$, the fraction is in lowest terms.
So $m = 71$, $n = 400$.

$$\dfrac{2556}{14400} = \dfrac{71}{400},\quad m = 71,\; n = 400$$

💡 Factor both numerator and denominator into primes — the common part is the GCD.

#3 Eliminate Possibilities 4.NBT.B.4 Step 7

Compute $m + n = 71 + 400 = 471$, matching choice (D).

$$m + n = 71 + 400 = 471 \Rightarrow \textbf{(D)}$$

💡 Final addition — matches the choice (D) exactly.

[1] #9 7.SP.C.7 Set up the sample space using symmetry. Each of the three people independently p

[2] #16 7.SP.C.8 Define event $A_i = $ "box $i$ ends up with three blocks of the same color" (for

[3] #7 7.SP.C.8 Compute $|A_{i_1} \cap \cdots \cap A_{i_k}|$ for any specific $k$-subset of boxe

[4] #2 7.SP.C.8 List all five terms (Tool #2). For $k = 1$: $S_1 = \binom{5}{1}(4!)^2 = 5 \cdot

[5] #16 6.NS.B.3 Apply PIE: $|A_1 \cup \cdots \cup A_5| = 2880 - 360 + 40 - 5 + 1 = 2556$. So the

[6] #7 6.NS.B.4 Reduce $\dfrac{2556}{14400}$ to lowest terms. $\gcd(2556, 14400)$: $2556 = 4 \cd

[7] #3 4.NBT.B.4 Compute $m + n = 71 + 400 = 471$, matching choice (D).

Review

Reasonableness: Sanity. Probability $\tfrac{71}{400} \approx 0.1775$, so about $18\%$ of arrangements have at least one same-color box. That feels right: for a single box, $P(A_i) = \tfrac{1}{5} \cdot \tfrac{1}{5} = \tfrac{1}{25} = 0.04$. With $5$ boxes, a naive overestimate (ignoring overlap) is $5 \cdot \tfrac{1}{25} = 0.20$, and the true value $0.1775$ is just slightly less — consistent with PIE removing the small overcount. Also $S_1 - S_2 = 2520$, so the first two PIE terms dominate; $S_3, S_4, S_5$ contribute $+36$ total — small corrections, as expected when individual events are rare.

Alternative: Tool #16 (Change Focus) more directly — count the complement: number of $(\text{Ben}, \text{Jasmin})$ pairs where NO box is uniform. Equivalently, for each box $i$, at least one of Ben/Jasmin disagrees with Ang. Use PIE on "box $i$ is matched" events again (same answer arises from $14400 - 2556 = 11844$ complementary outcomes). Or: more advanced — use derangement-like recurrence on the permutations agreeing nowhere with Ang, but PIE on the $5$ boxes is the cleanest direct path.

CCSS standards used (min grade 7)

4.NBT.B.4 Fluently add and subtract multi-digit whole numbers (Final $m + n = 71 + 400 = 471$.)
6.NS.B.3 Fluently add, subtract, multiply, and divide multi-digit decimals (Combining PIE terms $2880 - 360 + 40 - 5 + 1 = 2556$ and forming the ratio $2556/14400$.)
6.NS.B.4 Find greatest common factor and least common multiple of two numbers (Reducing $\tfrac{2556}{14400}$ to lowest terms by computing $\gcd = 36$.)
7.SP.C.7 Develop probability models and use them to find probabilities of events (Defining the uniform sample space of $(5!)^3$ outcomes and using symmetry to fix Ang's placement.)
7.SP.C.8 Find probabilities of compound events using organized lists, tables, and simulation (Applying inclusion-exclusion across the five "box $i$ is uniform" events to count favorable outcomes.)

⭐ 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: $\binom{5}{k}((5-k)!)^2$ ways. Alternate signs to get $2880 - 360 + 40 - 5 + 1 = 2556$ favorable out of $(5!)^2 = 14400$, reduce to $\tfrac{71}{400}$, and $m + n = 471$.

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: D

Review

CCSS standards used (min grade 7)

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