AMC 8 · 2004 · #2
Grade 3 countingProblem
How many different four-digit numbers can be formed by rearranging the four digits in 2004?
Pick an answer.
AMC 8 2004 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: Take the four digits of $2004$ — namely $2, 0, 0, 4$ — and rearrange them. How many different four-digit numbers can you make?
Givens: The four digits to rearrange are $2, 0, 0, 4$ (the digit $0$ appears twice); A four-digit number cannot start with $0$; Answer choices: (A) $4$, (B) $6$, (C) $16$, (D) $24$, (E) $81$
Unknowns: The number of distinct four-digit numbers formed by rearranging $2, 0, 0, 4$
Understand
Restated: Take the four digits of $2004$ — namely $2, 0, 0, 4$ — and rearrange them. How many different four-digit numbers can you make?
Givens: The four digits to rearrange are $2, 0, 0, 4$ (the digit $0$ appears twice); A four-digit number cannot start with $0$; Answer choices: (A) $4$, (B) $6$, (C) $16$, (D) $24$, (E) $81$
Plan
Primary tool: #2 Make a Systematic List
There are only four digits to shuffle, and the leading digit can only be $2$ or $4$. The whole answer space is tiny, so the cleanest approach is Tool #2 (Make a Systematic List): pick an ordering rule, write every valid number in that order, and count. No formulas needed — the list itself is the proof.
Execute — Answer: B
3.OA.A.3 Step 1 - Decide the ordering rule.
- List numbers from smallest to largest.
- The thousands digit must be $2$ or $4$ (it cannot be $0$, or the number would not be four digits), so first list every valid number that starts with $2$, then every one that starts with $4$.
💡 A clear rule before listing prevents missed cases and duplicates — the heart of Tool #2.
3.OA.A.3 Step 2 - List every valid number that starts with $2$.
- After placing $2$ in the thousands place, the leftover digits $\{0, 0, 4\}$ go in the last three slots.
- Order those three by where the $4$ sits (right end, middle, left end).
💡 Fixing the leading digit reduces the problem to placing one nonzero digit among two identical zeros — easy to enumerate.
3.OA.A.3 Step 3 - List every valid number that starts with $4$.
- Same idea: the leftover digits are $\{0, 0, 2\}$, and the $2$ can sit at the right end, the middle, or the left end of the remaining three slots.
💡 The structure mirrors the previous case — same shape, different leading digit.
1.OA.A.1 Step 4 - Combine the two lists and count.
- Three numbers start with $2$ and three start with $4$, with no overlap, so the total is $3 + 3$.
💡 The Addition Principle: when two case lists do not overlap, the total is the sum of their sizes.
3.OA.A.3 Decide the ordering rule. List numbers from smallest to largest. The thousands d 3.OA.A.3 List every valid number that starts with $2$. After placing $2$ in the thousands 3.OA.A.3 List every valid number that starts with $4$. Same idea: the leftover digits are 1.OA.A.1 Combine the two lists and count. Three numbers start with $2$ and three start wi Review
Reasonableness: Quick sanity check by a different count. All four digits can be ordered in $4! = 24$ ways, but the two $0$s are identical, so divide by $2! = 2$ to get $12$ distinct digit strings. Half of those start with $0$ (by symmetry — $0$ is just as likely to land in front as any specific digit when all $4$ positions are equally available; more precisely, of the $12$ strings, exactly $6$ start with $0$ since after fixing $0$ in front the remaining $\{0,2,4\}$ has $3! = 6$ arrangements). So the valid four-digit numbers number $12 - 6 = 6$, matching the list and answer (B).
Alternative: Tool #9 (Solve an Easier Related Problem): instead of listing, first ignore the "no leading zero" rule and count all distinguishable arrangements of $\{2,0,0,4\}$, which is $\dfrac{4!}{2!} = 6 \cdot 2 = 12$. Then subtract the bad ones that start with $0$: fixing $0$ in the thousands place leaves $\{0,2,4\}$ in the last three slots, giving $3! = 6$ arrangements. Valid total $= 12 - 6 = 6$, confirming (B).
CCSS standards used (min grade 3)
3.OA.A.3Use multiplication and division within 100 to solve word problems in situations involving equal groups, arrays, and measuring quantities (Treating each case (numbers starting with $2$, numbers starting with $4$) as an equal-sized group of $3$ arrangements, so the structure of the count is two equal groups.)1.OA.A.1Use addition and subtraction within 20 to solve word problems (Combining the two case totals with the Addition Principle: $3 + 3 = 6$.)
⭐ When the answer is a small number, just list them all in a sensible order. Here only $2$ or $4$ can lead, and each leading digit gives $3$ numbers — total $6$.
⭐ When the answer is a small number, just list them all in a sensible order. Here only $2$ or $4$ can lead, and each leading digit gives $3$ numbers — total $6$.
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