2021 AMC 10 B Problem 22

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To help you master 2021 AMC 10 B Problem 22, this page includes:

The original AMC problem to try, plus the correct answer

Clear, step-by-step solutions using different methods

Key ideas, useful formulas, and takeaways you can apply to similar problems

Related problems from past AMC exams, plus new practice problems that build the same skills

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2021 AMC10 B Problem 22

Solution

Solution 1

Inclusion-Exclusion Principle (PIE) on Probabilities

By fixing Ang’s block placement as the baseline, we can reframe the problem. For any box to get three identical blocks, Ben and Jasmin must independently match Ang’s color in that box. We can apply the Principle of Inclusion-Exclusion (PIE) to the five events representing a match in each box.

Principle of Inclusion-Exclusion: $P (E_{1} \cup \dots \cup E_{n}) = \sum P (E_{i}) - \sum P (E_{i} \cap E_{j}) + \sum P (E_{i} \cap E_{j} \cap E_{k}) - \dots \pm P (E_{1} \cap \dots \cap E_{n})$

Step 1: Fix Ang’s placement as a reference

Since Ang, Ben, and Jasmin act independently, the probability of them matching only depends on their relative placements. We can fix Ang’s placement without loss of generality. Assume Ang places color $i$ into box $i$ for $i = 1, 2, 3, 4, 5$ .

Step 2: Define the target events

For box $i$ to have 3 blocks of the same color, Ben and Jasmin must both place color $i$ into box $i$ . Let $E_{i}$ be the event that both Ben and Jasmin match Ang on box $i$ . We want to find the probability of the union of these events: $P (E_{1} \cup E_{2} \cup E_{3} \cup E_{4} \cup E_{5})$ .

Step 3: Calculate intersection probabilities

To apply PIE, we need the probability that both Ben and Jasmin match Ang in a specific set of $k$ boxes. The probability that one person matches Ang in $k$ specific boxes is the number of ways to arrange the remaining $5 - k$ blocks divided by the $5!$ total permutations. Because Ben and Jasmin act independently, the probability that they BOTH match Ang in those $k$ boxes is the square of the individual probability.

P (one person matches k specific boxes) = \frac{( 5 - k )!}{5 !}

P (both match k specific boxes) = (\frac{( 5 - k )!}{5 !})^{2}

Step 4: Apply the Principle of Inclusion-Exclusion

There are $(k 5)$ ways to choose a subset of $k$ boxes. We multiply this by the intersection probability for each $k$ from 1 to 5 and alternate the signs.

P (E_{1} \cup \dots \cup E_{5}) = k = 1 \sum 5 (- 1)^{k - 1} (k 5) (\frac{( 5 - k )!}{5 !})^{2}

Step 5: Evaluate the terms

We calculate the term for each $k$ : For $k = 1$ : $(1 5) (\frac{24}{120})^{2} = 5 (\frac{1}{5})^{2} = \frac{1}{5}$ For $k = 2$ : $(2 5) (\frac{6}{120})^{2} = 10 (\frac{1}{20})^{2} = \frac{1}{40}$ For $k = 3$ : $(3 5) (\frac{2}{120})^{2} = 10 (\frac{1}{60})^{2} = \frac{1}{360}$ For $k = 4$ : $(4 5) (\frac{1}{120})^{2} = 5 (\frac{1}{120})^{2} = \frac{5}{14400}$ For $k = 5$ : $(5 5) (\frac{1}{120})^{2} = 1 (\frac{1}{120})^{2} = \frac{1}{14400}$

Step 6: Sum the terms to find the final probability

Combine the calculated terms with alternating signs. To add them, we find a common denominator of $14400$ .

P = \frac{1}{5} - \frac{1}{40} + \frac{1}{360} - \frac{5}{14400} + \frac{1}{14400}

P = \frac{2880}{14400} - \frac{360}{14400} + \frac{40}{14400} - \frac{5}{14400} + \frac{1}{14400}

P = \frac{2880 - 360 + 40 - 5 + 1}{14400} = \frac{2556}{14400}

Step 7: Simplify the fraction

Divide the numerator and denominator by their greatest common divisor. Both are divisible by $36$ .

P = \frac{2556 \div 36}{14400 \div 36} = \frac{71}{400}

Correct answer: D. $471$
The probability is 71/400. Since 71 is prime and doesn’t divide 400,

m = 71

and

n = 400

are relatively prime. Thus,

m + n = 71 + 400 = 471

. This matches answer choice D. (Notice that evaluating only the first two terms of PIE yields

\frac{7}{40} ⟹ 47

, which is a trap answer A!)

Solution 2

Conditioning on Ben’s Matches Using Derangements

Instead of applying a 2D Inclusion-Exclusion directly, we can look at Ben’s placement first. We group his outcomes by exactly how many boxes $k$ he matches with Ang. For each $k$ , we then calculate the probability that Jasmin matches Ang in at least one of those $k$ specific boxes.

Derangement Numbers: The number of permutations of $n$ elements with no fixed points is denoted $D_{n}$ . For small $n$ : $D_{0} = 1, D_{1} = 0, D_{2} = 1, D_{3} = 2, D_{4} = 9, D_{5} = 44$ . The number of permutations with exactly $k$ fixed points is $(k n) D_{n - k}$ .

Step 1: Determine the distribution of Ben’s matches

Fix Ang’s permutation. Let $k$ be the exact number of boxes where Ben matches Ang. The number of such permutations for Ben is $(k 5) D_{5 - k}$ . We divide by $120$ to find the probability $P (k)$ that Ben has exactly $k$ matches.

k = 0 : (0 5) D_{5} = 44 ⟹ P (0) = \frac{44}{120}

k = 1 : (1 5) D_{4} = 45 ⟹ P (1) = \frac{45}{120}

k = 2 : (2 5) D_{3} = 20 ⟹ P (2) = \frac{20}{120}

k = 3 : (3 5) D_{2} = 10 ⟹ P (3) = \frac{10}{120}

k = 4 : (4 5) D_{1} = 0 ⟹ P (4) = 0

k = 5 : (5 5) D_{0} = 1 ⟹ P (5) = \frac{1}{120}

Step 2: Find Jasmin’s probability for each $k$

If Ben matched Ang on a specific set of $k$ boxes, we need Jasmin to match Ang on at least one of those $k$ boxes. We use 1D Inclusion-Exclusion on Jasmin’s permutations to find this probability, $P (J_{k})$ .

P (J_{1}) = \frac{1}{5} = \frac{24}{120}

P (J_{2}) = 2 (\frac{1}{5}) - 1 (\frac{1}{20}) = \frac{7}{20} = \frac{42}{120}

P (J_{3}) = 3 (\frac{1}{5}) - 3 (\frac{1}{20}) + 1 (\frac{1}{60}) = \frac{7}{15} = \frac{56}{120}

P (J_{5}) = 1 - P (no matches) = 1 - \frac{44}{120} = \frac{76}{120}

Step 3: Apply the Law of Total Probability

Multiply the probability of Ben matching exactly $k$ boxes by the probability of Jasmin matching at least one of those $k$ boxes, and sum over all $k$ .

P_{total} = k = 1 \sum 5 P (k) \cdot P (J_{k})

P_{total} = (\frac{45}{120}) (\frac{24}{120}) + (\frac{20}{120}) (\frac{42}{120}) + (\frac{10}{120}) (\frac{56}{120}) + (\frac{1}{120}) (\frac{76}{120})

Step 4: Calculate the final sum

Evaluate the products and add them together.

P_{total} = \frac{1080 + 840 + 560 + 76}{14400} = \frac{2556}{14400} = \frac{71}{400}

Correct answer: D. $471$
The total probability is 71/400. Thus

m + n = 471

, confirming our previous result.

Adapted Problems to Practice

1 / 10

What Formulas Do You Need for This Problem?

Principle of Inclusion-Exclusion (PIE):

P (E_{1} \cup E_{2} \cup \dots \cup E_{n}) = i \sum P (E_{i}) - i < j \sum P (E_{i} \cap E_{j}) + \dots + (- 1)^{n - 1} P (E_{1} \cap \dots \cap E_{n})

$E_{i}$ are events. This is valid for any events, whether they are independent or not. We wanted the probability that AT LEAST ONE box met a condition, which translates mathematically to the union of multiple events.

Derangements:

A derangement is a permutation of elements with no fixed points. The number of derangements of $n$ items is $D_{n}$ . The first few are $D_{1} = 0, D_{2} = 1, D_{3} = 2, D_{4} = 9, D_{5} = 44$ .

$n$ is the number of elements being permuted. We used derangements to count exactly how many boxes Ben matched with Ang, effectively partitioning the sample space into distinct cases.

Independence of Events:

P (A \cap B) = P (A) \cdot P (B)

Valid when events $A$ and $B$ are independent. Even though permutations have internal dependencies (placing a block in box 1 affects box 2), the actions of different people (Ben and Jasmin) are independent. This allowed us to square the individual probability to find the probability of both people matching Ang.

Key Takeaway: Remember This Strategy for Next Time

When multiple agents independently assign items to identical slots.

Try: Fix one agent's assignment as the 'standard' or 'identity' permutation.

Why: Because the outcome depends only on relative matches between the agents and they act independently, assuming one agent acts in a specific ordered way preserves all the mathematical structure while vastly reducing the size of the sample space you have to think about.

When asking for the probability that 'at least one' item meets a complex matching condition.

Try: Use the Principle of Inclusion-Exclusion (PIE) directly on probabilities.

Why: Intersection events (e.g., matching on exactly boxes 1 and 2) are usually much easier to calculate than union events (matching on box 1 OR box 2). If agents are independent, you can easily square or cube the base intersection probabilities.

When a problem involves permutations matching a specific arrangement.

Try: Remember the derangement numbers (0, 1, 2, 9, 44) to quickly count outcomes with exactly $k$ matches.

Why: The number of permutations with exactly $k$ fixed points is always $(k n) D_{n - k}$ . This lets you break down complex permutation problems into straightforward, distinct cases.

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2021 AMC 10 B Problem 22

In This Page:

2021 AMC10 B Problem 22