Problem 20 · 2017 AMC 8
Hard
Counting & Probability
careful-counting
An integer between 1000 and 9999, inclusive, is chosen at random. What is the probability that it is an odd integer whose digits are all distinct?
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Answer: B — 56/225.
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Hint 1 of 2
Two restrictions fight over the same digits: 'odd' lives at the units, 'no leading zero' lives at the thousands. Fill the most constrained positions first so the constraints don't trip over each other.
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Hint 2 of 2
Place the units digit first (odd: 5 choices), then the thousands (no 0, and distinct from units), then the looser hundreds and tens. Filling tightest-first keeps each later count clean.
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Approach: fill the most-constrained positions first
- Total 4-digit integers from 1000 to 9999: 9000. These are equally likely, so probability = (favorable)/9000.
- Place the trickiest digit first — the units must be odd: {1,3,5,7,9}, 5 choices. Doing units before thousands is the key, because the leading-zero rule then has clean counts.
- Thousands: can't be 0 and can't repeat the units digit → 10 − 2 = 8 choices.
- Hundreds: any digit except the 2 already used → 8 choices. Tens: any except the 3 used → 7 choices.
- Favorable = 5 × 8 × 8 × 7 = 2240, so probability = 2240/9000 = 56/225.
- Why this transfers: in 'distinct digits with a position rule' counts, always assign the most-restricted slot first — tackling 0/leading and parity constraints early prevents double-counting headaches.
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