1. **Problem statement:** Rudolf has four digits that can form the number 2025. He wants to know how many different numbers greater than 2025 can be made using these digits. 2. **Digits involved:** The digits are 2, 0, 2, and 5. 3. **Total permutations:** Since there are repeated digits (two 2's), the total number of distinct permutations is given by the formula for permutations of multiset: $$\frac{4!}{2!} = \frac{24}{2} = 12$$ 4. **List all permutations:** The distinct numbers formed are: - 0225 (invalid as leading zero) - 0252 (invalid as leading zero) - 2025 - 2052 - 2205 - 2250 - 2502 - 2520 - 5022 - 5202 - 5220 5. **Exclude numbers with leading zero:** 0225 and 0252 are not valid four-digit numbers. 6. **Count numbers greater than 2025:** - 2025 (equal, not greater) - 2052 (greater) - 2205 (greater) - 2250 (greater) - 2502 (greater) - 2520 (greater) - 5022 (greater) - 5202 (greater) - 5220 (greater) There are 8 numbers greater than 2025. **Final answer:** Rudolf can make **8** different numbers greater than 2025 using the digits 2, 0, 2, and 5.