AMC 8 · 2020 · #7
Easy mode Grade 4Problem
Imagine writing a four-digit number, like 2347. We want to count special four-digit numbers that follow three rules.
Rule 1: The number is bigger than 2020 and smaller than 2400.
Rule 2: All four digits are different from each other.
Rule 3: The digits go in increasing order from left to right. (For example, in 2347 the digits are 2,3,4,7, and each one is bigger than the one before it.)
How many four-digit numbers follow all three rules?
Pick an answer.
AMC 8 2020 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: Count the four-digit integers strictly between $2020$ and $2400$ whose four digits are all different and appear in strictly increasing order from left to right (so each digit is smaller than the next one).
Givens: The integer $N$ satisfies $2020 < N < 2400$; $N$ has four digits, call them $a, b, c, d$ from left to right; All four digits are distinct; The digits are in strictly increasing order: $a < b < c < d$; Answer choices: (A) $9$, (B) $10$, (C) $15$, (D) $21$, (E) $28$
Unknowns: The total number of integers $\overline{abcd}$ that satisfy all of the conditions above
Understand
Restated: Count the four-digit integers strictly between $2020$ and $2400$ whose four digits are all different and appear in strictly increasing order from left to right (so each digit is smaller than the next one).
Givens: The integer $N$ satisfies $2020 < N < 2400$; $N$ has four digits, call them $a, b, c, d$ from left to right; All four digits are distinct; The digits are in strictly increasing order: $a < b < c < d$; Answer choices: (A) $9$, (B) $10$, (C) $15$, (D) $21$, (E) $28$
Plan
Primary tool: #2 Make a Systematic List
Secondary: #3 Eliminate Possibilities
The problem says 'how many integers', so reach for Tool #2 (Systematic List). But before listing four-digit numbers blindly, use Tool #3 (Eliminate Possibilities) on the two leading digits: the range $2020 < N < 2400$ together with the strict-increase rule forces $a$ and $b$ to specific values, leaving only the last two digits free. Once $a$ and $b$ are pinned down, we just list all increasing pairs $(c, d)$ chosen from the digits larger than $b$ — a short, fully bounded list.
Execute — Answer: C
4.NBT.A.2 Step 1 - Pin down the thousands digit $a$.
- Since $N$ is between $2020$ and $2400$, $N$ is a four-digit number starting with $2$.
- (Anything starting with $1$ is at most $1999 < 2020$; anything starting with $3$ or more is at least $3000 > 2400$.) So $a = 2$.
💡 Comparing four-digit numbers by their leading digit is exactly Grade 4 place value: only numbers starting with $2$ land in the $2{,}xxx$ family.
4.NBT.A.2 Step 2 - Pin down the hundreds digit $b$.
- Because $a < b$ and $a = 2$, we need $b \geq 3$.
- Could $b$ be $4$ or larger?
- If $b = 4$, then the smallest legal $N$ would be $2456$ (with the smallest increasing $c, d$ being $5$ and $6$), and $2456 > 2400$.
- So $b = 4$ is too big, and any larger $b$ is worse.
- The only value left is $b = 3$.
💡 Squeezing $b$ between two inequalities is the same multi-digit comparison move, just applied to the hundreds place.
4.OA.A.3 Step 3 - Reduce the problem.
- Every valid $N$ now looks like $23cd$ with $3 < c < d \leq 9$.
- So we just need to count pairs $(c, d)$ chosen from $\{4, 5, 6, 7, 8, 9\}$ with $c < d$.
💡 Stripping the problem down to 'count pairs from a small set' is a Grade 4 multi-step word-problem move.
2.OA.C.4 Step 4 - List the pairs $(c, d)$ systematically, sorted by the smaller digit $c$.
- For each choice of $c$, $d$ is any larger digit in the set.
- Count the pairs in each group: $c=4$: $(4,5),(4,6),(4,7),(4,8),(4,9)$ — $5$ pairs.
- $c=5$: $(5,6),(5,7),(5,8),(5,9)$ — $4$ pairs.
- $c=6$: $3$ pairs.
- $c=7$: $2$ pairs.
- $c=8$: $1$ pair.
- $c=9$: $0$ pairs (no larger digit).
💡 Counting items grouped into rows of $5, 4, 3, 2, 1$ is the same as a Grade 2 rectangular-array total: just add the row sizes.
4.OA.A.3 Step 5 Each valid $(c, d)$ pair gives exactly one integer $\overline{23cd}$, so the count of integers equals the count of pairs.
💡 One-to-one correspondence between pairs and numbers means the totals match — Grade 4 multi-step reasoning.
4.NBT.A.2 Pin down the thousands digit $a$. Since $N$ is between $2020$ and $2400$, $N$ is 4.NBT.A.2 Pin down the hundreds digit $b$. Because $a < b$ and $a = 2$, we need $b \geq 3$ 4.OA.A.3 Reduce the problem. Every valid $N$ now looks like $23cd$ with $3 < c < d \leq 9 2.OA.C.4 List the pairs $(c, d)$ systematically, sorted by the smaller digit $c$. For eac 4.OA.A.3 Each valid $(c, d)$ pair gives exactly one integer $\overline{23cd}$, so the cou Review
Reasonableness: Spot-check three of the $15$ integers: $2345$ (digits $2{,}3{,}4{,}5$, strictly increasing, in range) — valid. $2378$ (digits $2{,}3{,}7{,}8$) — valid. $2389$ (digits $2{,}3{,}8{,}9$) — valid. The largest one we built is $2389 < 2400$, and the smallest is $2345 > 2020$, so the entire family sits inside the required range. The count $15$ matches answer (C), and the other choices ($9, 10, 21, 28$) don't arise from any natural mis-count here.
Alternative: Tool #9 (Solve an Easier Related Problem) followed by combinations: once you see that any valid $N$ is determined by choosing $2$ digits from the $6$-element set $\{4,5,6,7,8,9\}$ (the larger digit becomes $d$, the smaller becomes $c$ — only one ordering is increasing), the count is $\binom{6}{2} = \frac{6 \cdot 5}{2} = 15$. Same answer, but uses Grade-7-level formal combinatorics; the systematic-list approach above gets there with only Grade 4 tools.
CCSS standards used (min grade 4)
4.NBT.A.2Read and write multi-digit whole numbers and compare using symbols (Using place-value comparison on four-digit numbers to force $a = 2$ (from $2020 < N < 2400$) and then $b = 3$ (since $b \geq 4$ would push $N$ above $2400$).)4.OA.A.3Solve multi-step word problems using four operations with whole numbers (Reducing the original counting question to the sub-question 'count increasing pairs $(c, d)$ from $\{4,\dots,9\}$' and recognizing each pair maps to exactly one valid integer $\overline{23cd}$.)2.OA.C.4Use addition to find the total number of objects in rectangular arrays (Adding the row sizes $5 + 4 + 3 + 2 + 1 = 15$ when the increasing pairs are grouped by the smaller digit $c$.)
⭐ This AMC 8 problem only needs Grade 4 place-value comparison plus a tidy list you can add up — no fancy combinations formula required!
⭐ This AMC 8 problem only needs Grade 4 place-value comparison plus a tidy list you can add up — no fancy combinations formula required!
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