(c) 25
A number between 4000 and 5000 will have a 4 in the thousand’s place and since the number has to be divisible by 5, it will have 5 at unit’s place. The hundreds’ and tens’ place each can be filled with any of the five digits. So,
Total number of required numbers = 1 × 5 × 5 × 1 = 25.