AMC 8 · 2015 · #10
Grade 5 countingProblem
How many integers between 1000 and 9999 have four distinct digits?
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: Count how many integers from $1000$ to $9999$ use four different digits — that is, no digit appears twice in the number.
Givens: The integer is between $1000$ and $9999$, so it has exactly $4$ digits; Available digits for each position are $\{0, 1, 2, 3, 4, 5, 6, 7, 8, 9\}$ ($10$ digits total); All four digits in the number must be distinct (no repeats); Answer choices: (A) $3024$, (B) $4536$, (C) $5040$, (D) $6480$, (E) $6561$
Unknowns: The total count of $4$-digit integers whose four digits are all different
Understand
Restated: Count how many integers from $1000$ to $9999$ use four different digits — that is, no digit appears twice in the number.
Givens: The integer is between $1000$ and $9999$, so it has exactly $4$ digits; Available digits for each position are $\{0, 1, 2, 3, 4, 5, 6, 7, 8, 9\}$ ($10$ digits total); All four digits in the number must be distinct (no repeats); Answer choices: (A) $3024$, (B) $4536$, (C) $5040$, (D) $6480$, (E) $6561$
Plan
Primary tool: #7 Identify Subproblems
Secondary: #9 Solve an Easier Related Problem
A $4$-digit number is just four slots to fill — thousands, hundreds, tens, units. Tool #7 (Identify Subproblems) splits the count into four independent choices, one per slot, so the multiplication principle gives the total. To convince ourselves the multiplication is right, Tool #9 (Solve an Easier Related Problem) starts with a $2$-digit version we can list by hand, then extends the pattern up to $4$ digits.
Execute — Answer: B
3.OA.A.1 Step 1 - Warm up with a smaller version.
- How many $2$-digit numbers ($10$-$99$) have two different digits?
- Thousands of cases is too many to list, but two-digit is doable.
- The tens digit has $9$ choices ($1$-$9$, no leading $0$).
- The units digit has $9$ choices ($0$-$9$ minus the one already used).
- That gives $9 \times 9 = 81$ such numbers.
- Spot-check: from $10$ to $19$, the only repeat is $11$, leaving $9$ — matches the $9$ choices for the units digit when the tens digit is fixed.
💡 Solving the easier $2$-digit case first shows that "choices per slot, then multiply" really works.
4.OA.A.3 Step 2 - Slot $1$ — thousands digit.
- The number must be at least $1000$, so the thousands digit cannot be $0$.
- That leaves $9$ allowed digits: $1, 2, 3, 4, 5, 6, 7, 8, 9$.
💡 Treating one digit position as its own mini-problem with its own constraint is the Tool #7 subproblems move.
4.OA.A.3 Step 3 - Slot $2$ — hundreds digit.
- Any of the $10$ digits $0$-$9$ is allowed, but it must be different from the thousands digit.
- That removes exactly $1$ option, leaving $10 - 1 = 9$ choices.
- Note the hundreds digit *is* allowed to be $0$ — only the leading digit has the no-zero rule.
💡 Each slot's count depends on how many digits are still unused; subtract used digits from $10$.
4.OA.A.3 Step 4 - Slot $3$ — tens digit.
- It must differ from both the thousands digit and the hundreds digit, so $2$ digits are off-limits.
- That leaves $10 - 2 = 8$ choices.
💡 Same pattern: $10$ digits minus those already used.
4.OA.A.3 Step 5 - Slot $4$ — units digit.
- It must differ from the three digits already used, leaving $10 - 3 = 7$ choices.
💡 Last slot has the fewest options because the most digits are already taken.
5.OA.A.1 Step 6 - Multiply the per-slot choices together — this is the multiplication principle: independent choices combine by multiplying.
- Compute $9 \times 9 = 81$, then $8 \times 7 = 56$, then $81 \times 56 = 4536$.
💡 Each combination of slot choices gives a different number, and every valid number arises this way exactly once.
3.OA.A.1 Warm up with a smaller version. How many $2$-digit numbers ($10$-$99$) have two 4.OA.A.3 Slot $1$ — thousands digit. The number must be at least $1000$, so the thousands 4.OA.A.3 Slot $2$ — hundreds digit. Any of the $10$ digits $0$-$9$ is allowed, but it mus 4.OA.A.3 Slot $3$ — tens digit. It must differ from both the thousands digit and the hund 4.OA.A.3 Slot $4$ — units digit. It must differ from the three digits already used, leavi 5.OA.A.1 Multiply the per-slot choices together — this is the multiplication principle: i Review
Reasonableness: Sanity check: the total count of all $4$-digit numbers (with no distinctness rule) is $9 \times 10 \times 10 \times 10 = 9000$. Our answer $4536$ is about half of $9000$, which feels right — distinct-digit numbers should be a sizable but not majority share. The units-digit trick also confirms it: $9 \times 9 \times 8 \times 7$ ends in $1 \times 6 = 6$, and only choice (B) $4536$ ends in $6$. Choices (A) $3024$ and (C) $5040$ are $9 \times 8 \times 7 \times 6$ and $10 \times 9 \times 8 \times 7$ — common traps that forget the leading-zero rule or apply it twice.
Alternative: Tool #16 (Count the Complement) is harder here because "$4$-digit numbers with at least one repeated digit" splits into messy cases (exactly one pair, two pairs, three of a kind, four of a kind). A cleaner alternative is Tool #2 (Make a Systematic List) on a tiny version: list all $2$-digit numbers with distinct digits to confirm $81$, then trust the same per-slot logic scales to $4$ slots — exactly the bridge our easier-problem warm-up provided.
CCSS standards used (min grade 5)
3.OA.A.1Interpret products of whole numbers (Reading $9 \times 9$ as "$9$ groups of $9$" when counting two-digit distinct-digit numbers in the warm-up — the foundation for the multiplication principle.)4.OA.A.3Solve multistep word problems using the four operations (Computing each slot's choice count by subtracting already-used digits from $10$ ($10 - 1$, $10 - 2$, $10 - 3$) — a multistep reasoning move inside the larger counting problem.)5.OA.A.1Use parentheses, brackets, or braces in numerical expressions, and evaluate expressions with these symbols (Evaluating the chained product $9 \times 9 \times 8 \times 7$ by grouping as $(9 \times 9) \times (8 \times 7) = 81 \times 56 = 4536$.)
⭐ Big counting problems get easy when you split them into one small choice per slot, then multiply — a Grade 5 expression-evaluation move.
⭐ Big counting problems get easy when you split them into one small choice per slot, then multiply — a Grade 5 expression-evaluation move.
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