🇺🇸 AMC 8 ⇄ switch contest
2015 AMC 8

Problem 10

Problem 10 · 2015 AMC 8 Easy
Counting & Probability careful-counting

How many integers between 1000 and 9999 have four distinct digits?

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Answer: B — 4536 integers.
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Hint 1 of 2
Build the 4-digit number one place at a time, left to right. 'Distinct' means every place must dodge the digits already used — so the menu of choices shrinks by one each step.
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Hint 2 of 2
Handle the trickiest restriction first: the leading digit can't be 0 (else it isn't a 4-digit number). Then multiply the choices for the four places.
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Approach: fill the places left to right; multiply the shrinking choice counts
  1. Thousands place: 9 choices (1–9; 0 is banned as a leading digit).
  2. Hundreds place: now 0 is allowed again, but the thousands digit is used up — that's 10 − 1 = 9 choices (the count stays 9, just for a different reason).
  3. Tens: 8 left. Ones: 7 left.
  4. Total = 9 × 9 × 8 × 7 = 4536.
  5. Why this transfers: tackle the most-restricted slot first, then count the rest as 'whatever's left.' Watch the subtle coincidence here — the hundreds place is 9 because a banned-zero rule and a used-up-digit rule happen to remove one option each.
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