Problem 10 · 2012 AMC 8
Medium
Counting & Probability
permutations-with-repeats
How many 4-digit numbers greater than 1000 are there that use the four digits of 2012?
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Answer: D — 9.
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Hint 1 of 2
Two of the four digits are identical (two 2's), and the 0 can't lead. The cleanest plan: handle the picky positions first — decide where the 0 goes, then where the 1 goes, and the two 2's fill the rest with no further choice.
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Hint 2 of 2
This is place the constrained items first. Because the two 2's are interchangeable, once 0 and 1 are placed there's exactly one way to finish — so just multiply the number of choices.
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Approach: place the 0 (restricted) and the 1, the 2's fall in automatically
- The digits are {0, 1, 2, 2}. The 0 is the troublemaker — it can't be the leading digit, so it has only 3 legal spots (not 4).
- After the 0 is down, the 1 takes any of the 3 remaining spots.
- The last two spots must both be 2's — identical, so there's nothing left to decide. Total = 3 × 3 = 9 numbers.
- The reusable idea: when something is restricted (a digit that can't lead, a person who must sit at the end), place it first — you count the easy cases and never over- or under-count.
Another way — all arrangements, then remove the bad ones:
- Arrange 4 digits where two are identical: 4! / 2! = 12 total orderings.
- Toss out the ones that start with 0: fix 0 in front and arrange {1, 2, 2}, giving 3! / 2! = 3 bad cases.
- Good numbers = 12 − 3 = 9. (The 2! divides out the swap of the two identical 2's.)
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