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, 2022

Combinatorics Difficulty 5.0 AIME, harder Prove it United States

Problem:
A number is chosen uniformly at random from the set of all positive integers with at least two digits, none of which are repeated. Find the probability that the number is even.

Solution

Solution:
Since the number has at least two digits, all possible combinations of first and last digits have the same number of possibilities, which is $\sum_{i=0}^{8} \frac{8!}{i!}$ . Since the first digit cannot be zero, all of the last digits have 8 possible first digits, except for 0, which has 9 possible first digits. Therefore, the probability that the last digit is even is $\frac{9+4 \cdot 8}{9+9 \cdot 8}=\frac{41}{81}$ .

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