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Harvard-MIT Math Tournament

Combinatorics Difficulty 4.7 AIME Prove it United States

Problem:

How many four-digit numbers are there in which at least one digit occurs more than once?

Solution

Solution:

There are $9000$ four-digit numbers altogether. If we consider how many four-digit numbers have all their digits distinct, there are $9$ choices for the first digit (since we exclude leading zeroes), and then $9$ remaining choices for the second digit, then $8$ for the third, and $7$ for the fourth, for a total of $9 \cdot 9 \cdot 8 \cdot 7 = 4536$ . Thus the remaining $9000 - 4536 = 4464$ numbers have a repeated digit.

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.