To solve this problem, we first list the available digits: $1, 7, 8, 9$. These are the only digits we can use to form the 3-digit numbers.

For a valid 3-digit number, the first digit cannot be 0 (though 0 is excluded in this case), but any of the digits 1, 7, 8, 9 are allowed for the first digit. So, for the first digit, there are 4 choices.

Similarly, for the second and third digits, since no restrictions are given other than the exclusion of 0, 2, 3, 4, 5, and 6, we can again choose from the digits 1, 7, 8, 9. Therefore, there are 4 choices for both the second and third digits.

The total number of 3-digit numbers is the product of the choices for each digit: $4 \times 4 \times 4 = 4^3 = 64$

Thus, 64 three-digit numbers can be formed using the digits 1, 7, 8, and 9.

---

Would you like more details or examples? Here are some related questions:

1. How many 3-digit numbers can be formed if no digits are repeated?
2. What if we allow the use of 0, but not as the first digit?
3. How many 4-digit numbers can be formed with the same restrictions?
4. What is the probability of randomly forming a 3-digit number where all digits are the same?
5. How many 3-digit numbers can be formed if digits can repeat, but 9 must always be included?

Tip: When forming numbers with restrictions, always start by identifying how many choices are available for each digit position.