AMC 8 · 2015 · #10
Easy mode Grade 5Problem
Think about all the 4-digit whole numbers — every number from 1000 to 9999.
We want to count only the ones where all four digits are different from each other. For example, 1234 counts (all four digits are different), but 1123 does not count (the digit 1 shows up twice).
How many 4-digit numbers have four different 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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