1. The problem asks: How many 4-digit numbers greater than 4000 can be formed using the digits 2, 3, 4, 5, 6, and 7, with all digits different? 2. To solve this, we use the counting principle and permutations. 3. Since the number must be greater than 4000, the first digit can be 4, 5, 6, or 7 (4 choices). 4. After choosing the first digit, we have 5 remaining digits to fill the next 3 places, all different. 5. The number of ways to arrange the remaining 3 digits from 5 digits is a permutation: $P(5,3) = \frac{5!}{(5-3)!} = 5 \times 4 \times 3 = 60$. 6. Total numbers = choices for first digit $\times$ permutations of remaining digits = $4 \times 60 = 240$. 7. Therefore, the answer is 240. Final answer: 240