Problem 8 · AMC 8 Stretch
Core
Counting & Probability
accounting-for-all-possibilitiessymmetry
A 'symmetrical' plate has the form \(N_1 N_2 N_1 - L_1 L_2 L_1\): the first and third digits match, the first and third letters match, and the middle entry is different from the outer ones (like 363-WXW). Out of \(17{,}576{,}000\) total plates, how many are symmetrical? What is the probability of getting one?
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Answer: 58,500 plates; probability about 0.00333 (about 1 in 300)
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Hint 1 of 3
Symmetry forces the third digit to copy the first and the third letter to copy the first. Those positions are not free choices.
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Hint 2 of 3
Pick the first digit (10 ways), then the middle digit different from it (9 ways); the third digit is forced (1 way). Do the same for letters with 26 and 25.
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Hint 3 of 3
Multiply the digit count by the letter count, then divide by \(17{,}576{,}000\).
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Approach: Multiplication principle with forced positions
- Build the symmetrical digit part \(N_1 N_2 N_1\): choose \(N_1\) in 10 ways, choose \(N_2\) different from it in 9 ways, and \(N_3\) is forced to equal \(N_1\) (1 way): \(10\times 9\times 1=90\).
- Same for the letters \(L_1 L_2 L_1\): \(26\times 25\times 1=650\).
- So the number of symmetrical plates is \(90\times 650=58{,}500\), and the probability is \(\frac{58{,}500}{17{,}576{,}000}\approx 0.00333\), about 1 in 300.
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