AMC 8 · 2015 · #11
Easy mode Grade 7Problem
In the small country of Mathland, every license plate has 4 symbols in a row. The rules for the symbols are:
- The 1st symbol must be a vowel: one of A, E, I, O, U (5 choices).
- The 2nd symbol must be a non-vowel letter (21 choices).
- The 3rd symbol must also be a non-vowel letter, and it must be different from the 2nd symbol.
- The 4th symbol must be a digit from 0 to 9 (10 choices).
A license plate is made by picking each symbol at random, following these rules.
What is the probability that the plate comes out as exactly "AMC8"?
Pick an answer.
AMC 8 2015 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: 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"?
Givens: Position $1$: one of $5$ vowels (A, E, I, O, U); Position $2$: one of $21$ non-vowel letters; Position $3$: one of the $21$ non-vowels, but different from position $2$; Position $4$: one of $10$ digits ($0$-$9$); All legal plates are equally likely; Answer choices: (A) $\tfrac{1}{22{,}050}$, (B) $\tfrac{1}{21{,}000}$, (C) $\tfrac{1}{10{,}500}$, (D) $\tfrac{1}{2{,}100}$, (E) $\tfrac{1}{1{,}050}$
Unknowns: The probability that a random legal plate is exactly "AMC8"
Understand
Restated: 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"?
Givens: Position $1$: one of $5$ vowels (A, E, I, O, U); Position $2$: one of $21$ non-vowel letters; Position $3$: one of the $21$ non-vowels, but different from position $2$; Position $4$: one of $10$ digits ($0$-$9$); All legal plates are equally likely; Answer choices: (A) $\tfrac{1}{22{,}050}$, (B) $\tfrac{1}{21{,}000}$, (C) $\tfrac{1}{10{,}500}$, (D) $\tfrac{1}{2{,}100}$, (E) $\tfrac{1}{1{,}050}$
Plan
Primary tool: #7 Identify Subproblems
Secondary: #2 Make a Systematic List
Probability here is $\dfrac{1}{\text{total legal plates}}$, 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.
Execute — Answer: B
7.SP.C.8 Step 1 - Subproblem 1 — count the choices for position $1$.
- The vowels are listed: A, E, I, O, U.
- That is $5$ letters.
💡 Listing the sample space for one slot is a Grade 7 "organized list" move for compound events.
4.OA.A.3 Step 2 - Subproblem 2 — count the choices for position $2$.
- The English alphabet has $26$ letters and $5$ of them are vowels, so the non-vowels number $26 - 5 = 21$.
- Any of those $21$ may go in position $2$.
💡 Subtracting the vowel count from $26$ is a Grade 4 multi-step subtraction inside a counting word problem.
7.SP.C.8 Step 3 - Subproblem 3 — count the choices for position $3$.
- It must be a non-vowel different from position $2$, so one of the $21$ non-vowels is already used up.
💡 Treating position $3$ as a dependent slot (one option removed) is exactly the "compound event with the previous outcome subtracted" idea.
7.SP.C.8 Step 4 - Subproblem 4 — count the choices for position $4$.
- The digits $0, 1, 2, 3, 4, 5, 6, 7, 8, 9$ give $10$ options.
💡 Another organized-list count of the per-slot sample space.
4.OA.A.3 Step 5 - Combine the four subproblems by the multiplication principle to get the total number of legal plates.
- Each plate is one independent pick from each slot, so the counts multiply.
💡 Combining the per-slot counts by multiplication is the Grade 4 multi-step "how many in total" pattern.
7.SP.C.7 Step 6 - Check that "AMC8" is itself a legal plate: A is a vowel (✓ pos $1$), M is a non-vowel (✓ pos $2$), C is a different non-vowel (✓ pos $3$), $8$ is a digit (✓ pos $4$).
- So "AMC8" is one specific outcome out of the $21{,}000$ equally-likely legal plates, giving probability $\tfrac{1}{21{,}000}$.
💡 Forming probability as "favorable / total" on an equally-likely sample space is the Grade 7 probability-model definition.
7.SP.C.8 Subproblem 1 — count the choices for position $1$. The vowels are listed: A, E, 4.OA.A.3 Subproblem 2 — count the choices for position $2$. The English alphabet has $26$ 7.SP.C.8 Subproblem 3 — count the choices for position $3$. It must be a non-vowel differ 7.SP.C.8 Subproblem 4 — count the choices for position $4$. The digits $0, 1, 2, 3, 4, 5, 4.OA.A.3 Combine the four subproblems by the multiplication principle to get the total nu 7.SP.C.7 Check that "AMC8" is itself a legal plate: A is a vowel (✓ pos $1$), M is a non- Review
Reasonableness: The denominator must equal the total number of legal plates, and $5 \times 21 \times 20 \times 10$ visibly equals $21{,}000$. Choice (A) $22{,}050 = 5 \times 21 \times 21 \times 10$ 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".
Alternative: Tool #3 (Eliminate Possibilities) on the choices: rewrite each denominator as a product of factors that fit "vowels $\times$ consonants $\times$ consonants $\times$ digits". $22{,}050 = 5 \cdot 21 \cdot 21 \cdot 10$ assumes positions $2$ and $3$ may repeat — eliminated by the "different" constraint. $10{,}500 = 5 \cdot 21 \cdot 20 \cdot 5$ halves the digit count, which the problem does not. $2{,}100$ and $1{,}050$ drop another factor entirely. Only $21{,}000 = 5 \cdot 21 \cdot 20 \cdot 10$ matches every constraint, so (B) survives.
CCSS standards used (min grade 7)
4.OA.A.3Solve multi-step word problems using four operations with whole numbers (Subtracting $26 - 5 = 21$ for the non-vowel count and multiplying the per-slot counts $5 \times 21 \times 20 \times 10 = 21{,}000$ for the total.)7.SP.C.7Develop probability models and use them to find probabilities of events (Treating every legal plate as equally likely and reading the probability of "AMC8" as $\tfrac{1}{\text{total plates}}$.)7.SP.C.8Find probabilities of compound events using organized lists, tables, and simulation (Counting the sample space slot-by-slot ($5$ vowels, $21$ non-vowels, $20$ remaining non-vowels, $10$ digits) and combining by the multiplication principle for compound events.)
⭐ 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.
⭐ 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.
More like this
Same archetype — closest grade level first.