How many distinct 4-digit numbers can be formed using the digits 1, 2, 3, 4, 5, 6 (each digit can be used at MOST once)?

1. Each 4-digit number is an ORDERED selection of 4 from 6 distinct digits (no repetition). 2. This is a PERMUTATION: $P(6, 4) = \dfrac{6!}{(6-4)!} = \dfrac{6!}{2!} = \dfrac{720}{2} = 360$. 3. Equivalently: 6 choices for first digit, 5 for second, 4 for third, 3 for fourth $= 6 \cdot 5 \cdot 4 \cdot 3 = 360$. 4. Option A is $6!$ which uses ALL 6 digits. Option C is the unordered selection — doesn't account for digit position. Option D allows REPETITION (which we excluded). _Source: syllabus Mathematics, Ch 6, §6.3 + Examples, p. 6–7._