Bifalgorithm post · Brisk · Chapter 1

How can the counting rule handle positions with different numbers of choices?

How can the counting rule handle positions with different numbers of choices? (Brisk · Chapter 1) — "Suppose Virginia outlawed personalized plates and gave eve…

Ideas in this flowchart

  • Restricting the last four positions of seven-character plates to digits greatly reduces the number of possible plates.
  • Virginia outlaws personalized plates and assigns each vehicle a randomly generated seven-character plate.
  • The last four characters must be digits, so a plate such as “RFP-6YQ7” is impossible.
  • Positions 1 through 3 each have 36 possible choices.
  • Positions 4 through 7 each have only 10 possible choices.
  • The total is calculated as 36 × 36 × 36 × 10 × 10 × 10 × 10.
  • This produces 466,560,000 possible plates.
  • The restricted system provides only about 0.006 times as many plates as the previous system.
  • All seven positions should use alphanumeric characters instead.
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How can the counting rule handle positions with different numbers of choices?

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Brisk

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"Suppose Virginia outlawed personalized plates and gave everyone a randomly generated 7-character plate. Furthermore, the last four characters of the plate had to be digits instead of letters, so that something like “RFP-6YQ7” would be impossible. In this case, not each of the k parts has an equal number of choices. n₁ through n₃ are still 36, but now n₄ through n₇ are just 10. So this gives us: 36 × 36 × 36 × 10 × 10 × 10 × 10 = 466,560,000 plates, or only about .006 times as many as before. Better stick with alphanumeric characters for all seven positions."

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