Problem 2 · 2004 AMC 8
Easy
Counting & Probability
permutations-with-repeats
How many different four-digit numbers can be formed by rearranging the four digits in 2004?
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Answer: B — 6.
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Hint 1 of 2
There are really only two 'interesting' digits here — the 2 and the 4. The two zeros are interchangeable, and a number can't start with 0. What position can each non-zero digit take?
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Hint 2 of 2
Whenever a digit repeats, ordinary 4! over-counts; and a 'no leading zero' rule trims more. Naming both traps — identical-item over-count and the leading-zero restriction — is the whole skill.
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Approach: place the non-zero digits first
- The slick view: a valid number is fixed once you decide where the 2 and 4 sit, because the leftover slots must be the two zeros. The thousands place must be 2 or 4 (no leading 0) — 2 choices — and the other non-zero digit then has 3 open slots — 3 choices. That's 2 × 3 = 6.
- Why this beats brute force: you never list anything; the two zeros take care of themselves once the 'real' digits are placed.
- You'll meet this again any time a number has repeated digits and a 'first digit can't be 0' rule — place the restricted/distinct items first, let the duplicates fall into the gaps.
Another way — count all, then subtract leading zeros:
- All arrangements of {2, 0, 0, 4}: 4!/2! = 12 (divide by 2! because the two 0s are identical).
- Bad ones put a 0 in front: the remaining three slots hold {2, 0, 4}, giving 3! = 6 arrangements.
- Valid = 12 − 6 = 6.
Another way — just list them:
- Smallest first: 2004, 2040, 2400, 4002, 4020, 4200.
- That's exactly 6 — a fine check when the count is this small.
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