To solve the problem:

We are tasked with finding the number of five-digit numbers with distinct digits that do not include the digits 4, 6, or 7.

Step-by-Step Solution:

1. Available digits: The total digits are $\{0, 1, 2, 3, 5, 8, 9\}$. These exclude $\{4, 6, 7\}$, leaving us with 7 digits.

2. Creating five-digit numbers:

   • A five-digit number cannot start with $0$ (as it would no longer be a five-digit number).
   • So, for the first digit, we have $6$ valid choices ($\{1, 2, 3, 5, 8, 9\}$).
   • For the second digit, we have $6$ remaining choices (since one digit is already used, and $0$ is now valid).
   • For the third digit, $5$ choices remain.
   • For the fourth digit, $4$ choices remain.
   • For the fifth digit, $3$ choices remain.

3. Calculation:

   • The total number of valid five-digit numbers is: $6 \times 6 \times 5 \times 4 \times 3 = 2160$

Final Answer:

There are 2160 five-digit numbers with distinct digits that do not contain $4$, $6$, or $7$.

Do you need further explanation or breakdowns?

Related Questions:

1. How would the calculation change if all digits were allowed?
2. How many such numbers would there be if repetition of digits were allowed?
3. What if the number must begin with an odd digit?
4. How many four-digit numbers can be formed under similar restrictions?
5. Can you derive a general formula for $n$-digit numbers under these constraints?

Tip:

When solving combinatorics problems, always identify constraints like digit exclusions and ordering restrictions to narrow the scope effectively.