1. The problem is to create a trio (group of three numbers) from the digits 1, 2, 3, 4, 5, 6 without repeating any sequence. 2. We interpret "without repeating a sequence" as forming unique 3-digit numbers where the order of digits matters and no digit repeats within the trio. 3. The formula for the number of permutations of $n$ distinct objects taken $r$ at a time is: $$P(n,r) = \frac{n!}{(n-r)!}$$ where $n=6$ (digits 1 to 6) and $r=3$ (trio length). 4. Calculate the number of unique 3-digit sequences: $$P(6,3) = \frac{6!}{(6-3)!} = \frac{6!}{3!} = \frac{720}{6} = 120$$ 5. This means there are 120 unique trios possible without repeating any digit in the sequence. 6. To create one example trio, select any 3 distinct digits, for example, 1, 2, and 3, and arrange them in any order, e.g., 123. 7. Thus, one valid trio is 123, and similarly, other unique trios can be formed without repeating sequences. Final answer: There are 120 unique trios possible from digits 1 to 6 without repeating any sequence.