AMC 8 · 2012 · #10

Easy mode Grade 7

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Problem

Take the four digits $2, 0, 1, 2$ . You can rearrange them to form $4$ -digit numbers, such as $2012$ or $1220$ .

A $4$ -digit number cannot start with $0$ , so it must be greater than $1000$ .

How many different $4$ -digit numbers can you make this way?

Pick an answer.

(A)

(B)

(C)

(D)

(E)

AMC 8 2012 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.

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Toolkit + CCSS Solution

Understand

Restated: Using exactly the four digits of $2012$ (that is, the multiset $\{0, 1, 2, 2\}$), how many $4$-digit numbers greater than $1000$ can we form?

Givens: The available digits are $\{0, 1, 2, 2\}$ — one $0$, one $1$, and two $2$s; All four digits must be used (each arrangement is a $4$-digit number); The number must be greater than $1000$; Answer choices: (A) $6$, (B) $7$, (C) $8$, (D) $9$, (E) $12$

Unknowns: The count of valid $4$-digit numbers

Understand

Restated: Using exactly the four digits of $2012$ (that is, the multiset $\{0, 1, 2, 2\}$), how many $4$-digit numbers greater than $1000$ can we form?

Plan

Primary tool: #4 Make a Systematic List

Secondary: #7 Identify Subproblems

There are only a handful of arrangements of $\{0, 1, 2, 2\}$, so Tool #4 (Make a Systematic List) is the safest way to count without missing or double-counting — especially since the two $2$s are identical. Tool #7 (Identify Subproblems) splits the list cleanly by the leading digit: either the number starts with $1$ or it starts with $2$ (it cannot start with $0$). Counting each case separately and adding is much easier than juggling the whole thing at once.

Execute — Answer: D

#7 Identify Subproblems 4.NBT.A.2 Step 1

Decide which digits can lead

Decide which digits can lead.
A $4$-digit number greater than $1000$ must have a nonzero thousands digit.
From $\{0, 1, 2, 2\}$ the nonzero options are $1$ and $2$, so split the count into two cases.

$$\text{leading digit} \in \{1, 2\}$$

💡 Knowing that the thousands place controls how big the number is — and that a $0$ there shrinks it to $3$ digits — is Grade 4 place-value reasoning.

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

List the numbers starting with 1

Case 1: the number starts with $1$.
The remaining digits to place are $\{0, 2, 2\}$.
List every order systematically: put $0$ first, second, or third among the remaining three slots.

$$1\underline{0}22,\ 1\underline{2}0\underline{2},\ 1\underline{2}\underline{2}0 \;\Rightarrow\; 3 \text{ numbers}$$

💡 Writing out the three positions for the lone $0$ is exactly the Grade 7 "organized list" method for counting outcomes.

#4 Make a Systematic List 7.SP.C.8 Step 3

List the numbers starting with 2

Case 2: the number starts with $2$.
The remaining digits to place are $\{0, 1, 2\}$, all different.
Systematically list every ordering of these three distinct digits.

$$2012,\ 2021,\ 2102,\ 2120,\ 2201,\ 2210 \;\Rightarrow\; 6 \text{ numbers}$$

💡 Three distinct items have $3 \times 2 \times 1 = 6$ orderings, and an organized list confirms every one.

#7 Identify Subproblems 4.OA.A.3 Step 4

Add the two cases

Add the two cases.
Every valid number lands in exactly one case (it starts with either $1$ or $2$, never both), so the cases don't overlap and the total is just the sum.

$$3 + 6 = 9 \;\Rightarrow\; \textbf{(D)}$$

💡 Combining disjoint case counts by addition is the Grade 4 multi-step word-problem move.

[1] #7 4.NBT.A.2 Decide which digits can lead. A $4$-digit number greater than $1000$ must have a

[2] #4 7.SP.C.8 Case 1: the number starts with $1$. The remaining digits to place are ${0, 2, 2

[3] #4 7.SP.C.8 Case 2: the number starts with $2$. The remaining digits to place are ${0, 1, 2

[4] #7 4.OA.A.3 Add the two cases. Every valid number lands in exactly one case (it starts with

Review

Reasonableness: Sanity check by a different count: arrangements of $\{0, 1, 2, 2\}$ ignoring the leading-digit rule total $\tfrac{4!}{2!} = 12$ (we divide by $2!$ because the two $2$s are interchangeable). The bad arrangements are the ones that start with $0$, which fix $0$ first and rearrange $\{1, 2, 2\}$ in the last three slots — that's $\tfrac{3!}{2!} = 3$. So the valid count is $12 - 3 = 9$, matching (D).

Alternative: Tool #3 (Find a Formula): use the multiset permutation formula directly. Total orderings of $\{0, 1, 2, 2\}$ are $\tfrac{4!}{2!} = 12$. Subtract the $\tfrac{3!}{2!} = 3$ orderings that begin with $0$ to get $12 - 3 = 9$. Same answer (D), no listing needed.

CCSS standards used (min grade 7)

4.NBT.A.2 Read, write, and compare multi-digit whole numbers using place value (Recognizing that the thousands digit cannot be $0$ for a true $4$-digit number greater than $1000$.)
4.OA.A.3 Solve multi-step word problems using the four operations (Adding the two disjoint case counts ($3 + 6 = 9$) to get the final total.)
7.SP.C.8 Find probabilities of compound events using organized lists, tables, tree diagrams, and simulation (Building an organized list of every arrangement of $\{0, 2, 2\}$ and of $\{0, 1, 2\}$ in each case to count without missing or double-counting.)

⭐ An organized list — split by what the leading digit can be — turns this AMC 8 counting problem into a quick Grade 7 sample-space exercise.

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: D

Decide which digits can lead

List the numbers starting with 1

List the numbers starting with 2

Add the two cases

Review

CCSS standards used (min grade 7)

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