AMC 8 · 2020 · #7

Grade 4 counting

Archetype Small-Domain Constrained Search

digit-constraintscombinations-basicsystematic-enumeration digit-constraintscaseworksystematic-enumeration ↑ Prerequisites: systematic-enumeration

📏 Medium solution 💡 3 insights

📘 View easy version →

Problem

How many integers between $2020$ and $2400$ have four distinct digits arranged in increasing order? (For example, $2347$ is one integer.).

$\textbf{(A) }\text{9} \qquad \textbf{(B) }\text{10} \qquad \textbf{(C) }\text{15} \qquad \textbf{(D) }\text{21}\qquad \textbf{(E) }\text{28}$

Pick an answer.

(A)

$text{9}$

(B)

$text{10}$

(C)

$text{15}$

(D)

$text{21}$

(E)

$text{28}$

View mode:

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

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

#3 Eliminate Possibilities 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 $a = 2$.

$$2020 < N < 2400 \;\Longrightarrow\; a = 2$$

💡 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.

#3 Eliminate Possibilities 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 $b = 3$.

$$a = 2,\; a < b,\; N < 2400 \;\Longrightarrow\; b = 3$$

💡 Squeezing $b$ between two inequalities is the same multi-digit comparison move, just applied to the hundreds place.

#2 Make a Systematic List 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 $c < d$.

$$N = \overline{23cd},\; c, d \in \{4, 5, 6, 7, 8, 9\},\; c < d$$

💡 Stripping the problem down to 'count pairs from a small set' is a Grade 4 multi-step word-problem move.

#2 Make a Systematic List 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).

$$5 + 4 + 3 + 2 + 1 + 0 = 15$$

💡 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.

#2 Make a Systematic List 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.

$$\text{count} = 15 \;\Longrightarrow\; \textbf{(C)}$$

💡 One-to-one correspondence between pairs and numbers means the totals match — Grade 4 multi-step reasoning.

[1] #3 4.NBT.A.2 Pin down the thousands digit $a$. Since $N$ is between $2020$ and $2400$, $N$ is

[2] #3 4.NBT.A.2 Pin down the hundreds digit $b$. Because $a < b$ and $a = 2$, we need $b \geq 3$

[3] #2 4.OA.A.3 Reduce the problem. Every valid $N$ now looks like $23cd$ with $3 < c < d \leq 9

[4] #2 2.OA.C.4 List the pairs $(c, d)$ systematically, sorted by the smaller digit $c$. For eac

[5] #2 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.2 Read 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.3 Solve 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.4 Use 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!

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: C

Review

CCSS standards used (min grade 4)

More like this