To solve the problem of finding $n$, the number of numbers divisible by 5 between 3000 and 6000 that have different digits, let's proceed step by step:

Step 1: Constraints on the number

We are considering numbers between 3000 and 6000, which means these are 4-digit numbers, starting with the digit 3, 4, or 5.

• The number must be divisible by 5, so the last digit must be either 0 or 5.

Step 2: Analyze the possible cases

We break this problem down into two cases based on the last digit:

Case 1: Last digit is 0

If the last digit is 0, the first digit must be 3, 4, or 5 (since the number is between 3000 and 6000). The second and third digits must be chosen from the remaining digits (0 is already used, and the first digit is used). All digits must be distinct.

• The first digit can be 3, 4, or 5 (3 choices).
• The second digit can be chosen from the remaining 8 digits (since 0 and the first digit are not allowed), giving 8 choices.
• The third digit can be chosen from the remaining 7 digits, giving 7 choices.

Thus, for this case, the number of such numbers is: $3 \times 8 \times 7 = 168$

Case 2: Last digit is 5

If the last digit is 5, the first digit must be 3 or 4 (since 5 cannot be the first digit, and the number must be between 3000 and 6000). The second and third digits must be chosen from the remaining digits (5 is already used, and the first digit is used). All digits must be distinct.

• The first digit can be 3 or 4 (2 choices).
• The second digit can be chosen from the remaining 8 digits (since 5 and the first digit are not allowed), giving 8 choices.
• The third digit can be chosen from the remaining 7 digits, giving 7 choices.

Thus, for this case, the number of such numbers is: $2 \times 8 \times 7 = 112$

Step 3: Total number of numbers divisible by 5

The total number of numbers divisible by 5 is the sum of the two cases: $n = 168 + 112 = 280$

Step 4: Find $\dfrac{n}{35}$

Now, we calculate: $\dfrac{n}{35} = \dfrac{280}{35} = 8$

Thus, the value of $\dfrac{n}{35}$ is $\boxed{8}$.

Would you like further details on any step or additional clarifications?

Here are 5 related questions to expand your understanding:

1. How would the solution change if the digits could be repeated?
2. What if the range was extended to numbers between 2000 and 6000?
3. How can we count the number of 4-digit numbers divisible by another number, say 3?
4. What if we wanted numbers divisible by both 5 and 2, and digits still had to be different?
5. How can we find the total number of distinct 4-digit numbers regardless of divisibility?

Tip: When counting permutations with restrictions, it's useful to break the problem into manageable cases, as done here with different last digits!