Problem 7 · AMC 8 Stretch
Core
Counting & Probability
accounting-for-all-possibilitiesorganizing-data
Using the same plate form \(N_1 N_2 N_3 - L_1 L_2 L_3\) (total \(17{,}576{,}000\) plates), how many plates have three EQUAL digits and three EQUAL letters (like 777-MMM)? What is the probability of getting one?
Show answer
Answer: 260 plates; probability about 0.0000148 (about 1 in 67,600)
Show hints
Hint 1 of 3
If all three digits must be equal, only the first digit is a real choice — the other two are forced to copy it.
Still stuck? Show hint 2 →
Hint 2 of 3
Three equal digits: \(10\times 1\times 1\). Three equal letters: \(26\times 1\times 1\). Multiply for the plate count.
Still stuck? Show hint 3 →
Hint 3 of 3
Divide by \(17{,}576{,}000\) to get the probability.
Show solution
Approach: Multiplication principle — forced positions count as 1 choice
- If all three digits are equal, the first digit can be any of 10, and the other two are forced to match it: \(10\times 1\times 1=10\) ways. Same idea for letters: \(26\times 1\times 1=26\) ways.
- Digits and letters are independent, so \(10\times 26=260\) plates have three equal digits and three equal letters.
- Probability: \(\frac{260}{17{,}576{,}000}\approx 0.0000148\), about 1 in 67,600.
Mark:
· log in to save