Problem 17 · 2016 AMC 8
Easy
Counting & Probability
complementary-counting
An ATM password at Fred's Bank is composed of four digits from 0 to 9, with repeated digits allowable. If no password may begin with the sequence 9, 1, 1, then how many passwords are possible?
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Answer: D — 9990 passwords.
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Hint 1 of 2
"How many are ALLOWED" is awkward to count head-on. But the FORBIDDEN ones are super easy to count — so count everything, then subtract the bad ones. (Total − bad = good.)
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Hint 2 of 2
A forbidden password has its first three digits locked as 9, 1, 1; only the fourth digit is free. So there are very few bad ones — count those, not the good ones.
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Approach: complementary counting (total minus forbidden)
- Count everything first: 4 digits, each 0–9, gives 104 = 10,000 possible passwords.
- Count the forbidden ones: they must read 9, 1, 1, _ — first three digits fixed, last digit any of 10. That's only 10 bad passwords.
- Good = total − bad = 10,000 − 10 = 9990.
- Why this transfers: when a rule says "everything EXCEPT a small forbidden set," counting the small forbidden set and subtracting is almost always faster than counting the survivors directly.
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