Problem 6 · AMC 8 Stretch
Core
Counting & Probability
accounting-for-all-possibilitiesorganizing-data
A plate has the form \(N_1 N_2 N_3 - L_1 L_2 L_3\): three digits (0-9) then three letters (A-Z). The total number of possible plates is \(10^3\times 26^3 = 17{,}576{,}000\). How many plates have all three digits different AND all three letters different? What is the probability of getting one?
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Answer: 11,232,000 plates; probability about 0.0639 (about 1 in 16)
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Hint 1 of 3
Handle the digit part and the letter part separately, then multiply, because the digits and letters are chosen independently.
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Hint 2 of 3
All-different digits: \(10\times 9\times 8\). All-different letters: \(26\times 25\times 24\) (each new letter avoids the ones already used).
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Hint 3 of 3
Multiply those two counts for the number of favorable plates, then divide by \(17{,}576{,}000\) for the probability.
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Approach: Multiplication principle — count digits and letters separately
- The digits and letters are picked independently, so count each part and multiply.
- Three different digits: \(10\times 9\times 8=720\). Three different letters: \(26\times 25\times 24=15{,}600\).
- Favorable plates: \(720\times 15{,}600=11{,}232{,}000\).
- Probability: \(\frac{11{,}232{,}000}{17{,}576{,}000}\approx 0.0639\), about 1 in 16.
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