To find the number of numbers that can be formed between 100 and 1000 using the digits 5, 6, 7, 8, and 9 without repetition, we can break down the problem into three parts:

1. Counting the number of choices for the hundreds digit
2. Counting the number of choices for the tens digit
3. Counting the number of choices for the units digit

Counting the number of choices for the hundreds digit:
Since the hundreds digit cannot be 0, we have 4 choices: 5, 6, 7, or 8.

Counting the number of choices for the tens digit:
Once we have chosen the hundreds digit, we can no longer use that digit. So, we have 4 remaining digits to choose from. Since the tens digit can be 0, we have 4 choices for this digit.

Counting the number of choices for the units digit:
Once we have chosen the hundreds and tens digits, we can no longer use those digits. So, we have 3 remaining digits to choose from. Since the units digit cannot be 0, we have 3 choices for this digit.

To find the total number of numbers that can be formed, we multiply the number of choices for each digit together:

Number of numbers = Number of choices for the hundreds digit * Number of choices for the tens digit * Number of choices for the units digit
= 4 * 4 * 3
= 48

However, we need to consider that we are counting numbers between 100 and 1000, inclusive. So, we need to subtract the number of numbers that are less than 100.

The numbers less than 100 are:
50, 51, 52, 53, 54, 57, 58, 59
60, 61, 62, 63, 64, 65, 67, 68, 69
70, 71, 72, 73, 74, 75, 76, 78, 79
80, 81, 82, 83, 84, 85, 86, 87, 89
90, 91, 92, 93, 94, 95, 96, 97, 98, 99

There are 40 numbers less than 100. Therefore, the total number of numbers that can be formed between 100 and 1000 is 48 - 40 = 8.

Hence, the correct answer is option D) 60.