AMC 8 · 2015 · #11

Grade 7 probability

Archetype Branching State Enumeration └ Sub-archetype Ordered Placement Count

probability-basicpermutations-basicsystematic-enumeration systematic-enumeration ↑ Prerequisites: fraction-arithmeticmulti-digit-arithmetic

📏 Short solution 💡 2 insights

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Problem

A Mathland license plate has 4 symbols: position 1 is a vowel (A, E, I, O, U), positions 2 and 3 are two different non-vowel letters (from the 21 consonants), and position 4 is a digit (0-9). If every legal plate is equally likely, what is the probability that a random plate spells exactly "AMC8"?

Pick an answer.

(A)

$frac{1}{22,050}$

(B)

$frac{1}{21,000}$

(C)

$frac{1}{10,500}$

(D)

$frac{1}{2,100}$

(E)

$frac{1}{1,050}$

AMC 8 2015 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Identify Subproblems

Probability here is 1/(total legal plates), so the real work is counting the total. Tool #7 (Subproblems) splits one big count into 4 tiny ones — "how many choices for each position?" — which then multiply. Tool #2 (Systematic List) is the bookkeeping behind each subproblem: list the vowels, the non-vowels, the leftover non-vowels, and the digits to make sure each count is honest. After multiplying, the probability falls out as 1 over that total because "AMC8" is exactly one of the equally-likely plates.

1STEP 1

Subproblem 1 — position 1 is a vowel: A, E, I, O, U gives 5 choices.

choices₁ = 5

2STEP 2

Subproblem 2 — position 2 is a non-vowel: 26 letters minus 5 vowels leaves 21 choices.

choices₂ = 26 - 5 = 21

3STEP 3

Subproblem 3 — position 3 is a non-vowel different from position 2, so one is used up: 20 choices.

choices₃ = 21 - 1 = 20

4STEP 4

Subproblem 4 — position 4 is a digit 0-9, giving 10 choices.

choices₄ = 10

5STEP 5

By the multiplication principle the total number of legal plates = 5 × 21 × 20 × 10 = 21,000.

total = 5 × 21 × 20 × 10 = (5 × 20) × (21 × 10) = 100 × 210 = 21,000

6STEP 6

"AMC8" is one valid plate — A vowel, M and C different non-vowels, 8 a digit — so P = $\frac{1}{21,000}$ → (B).

P(AMC8) =

\frac{1}{21,000}

→ (B)

Answer

\frac{1}{21,000}

The denominator must equal the total number of legal plates, and 5 × 21 × 20 × 10 visibly equals 21,000. Choice (A) 22,050 = 5 × 21 × 21 × 10 would be the count if positions 2 and 3 were allowed to repeat — but the problem forbids repeats, so (A) is the classic trap. Choices (C), (D), (E) shrink the denominator, which would only happen if the favorable count were larger than 1 — but "AMC8" is a single specific plate. So (B) is the only number consistent with both the constraints and "exactly one favorable plate".

💡Key takeaway

Count each slot's options separately, multiply them all to get the total, then put 1 over that — that's the Grade 7 way to handle a "random plate" probability.

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