1. **State the problem:** We need to find how many 3-digit numbers can be formed using the digits 2, 3, 4, 5, 6 without repeating any digit. 2. **Formula and rules:** When forming numbers without repetition, the number of possible arrangements is given by permutations. For $n$ distinct digits taken $r$ at a time, the number of permutations is $$P(n,r) = \frac{n!}{(n-r)!}$$ 3. **Apply the formula:** Here, $n=5$ (digits 2,3,4,5,6) and $r=3$ (3-digit numbers). 4. Calculate permutations: $$P(5,3) = \frac{5!}{(5-3)!} = \frac{5!}{2!} = \frac{120}{2} = 60$$ 5. **Interpretation:** There are 60 different 3-digit numbers that can be formed using the digits 2, 3, 4, 5, 6 without repetition. **Final answer:** 60