Problem 23 · 1995 AJHSME
Hard
Counting & Probability
multiplication-principle
How many four-digit whole numbers are there such that the leftmost digit is odd, the second digit is even, and all four digits are different?
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Answer: B — 1400.
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Hint 1 of 2
Build the number one digit-slot at a time and count the choices for each slot; the answer is the product. Fill the MOST restricted slots first (the odd and even ones) so the 'all different' rule is easy to track.
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Hint 2 of 2
Odd digits and even digits never overlap, so choosing the first (odd) and second (even) can never collide. Then each later digit just avoids the ones already placed.
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Approach: multiplication principle, filling the constrained slots first
- Handle the picky positions first. First digit must be odd → {1,3,5,7,9}, 5 choices. Second must be even → {0,2,4,6,8}, 5 choices. These can't clash, since no number is both odd and even — that's why we do them first.
- Now the 'all different' rule kicks in for the rest. Third digit: any of 10 digits except the 2 already used → 8 choices. Fourth: except the 3 used → 7 choices.
- Multiply the independent slot-counts: 5 × 5 × 8 × 7 = 1400.
- The technique: for 'how many arrangements' problems, count choices slot by slot and multiply — and always fill the tightest-constrained slots first so the leftover counts stay clean. Sanity check: ignoring repeats it'd be 5 × 5 × 10 × 10 = 2500, and removing the clashes drops it to 1400 — smaller, as it should be.
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