How many numbers divisible by 5 and lying between 3000 and 4000 can be formed by using the digits 3, 4, 5, 6, 7, 8 when no digit is repeated in any such number?

Explanation

To solve the problem of how many numbers divisible by 5 and lying between 3000 and 4000 can be formed using the digits 3, 4, 5, 6, 7, and 8 without repeating any digit, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Range and Conditions**:
- The numbers must be between 3000 and 4000.
- Therefore, the first digit must be 3.
- The number must be divisible by 5, which means the last digit must be either 0 or 5. Since 0 is not among the available digits, the last digit must be 5.

2. **Fix the First and Last Digits**:
- First digit (thousands place): 3
- Last digit (units place): 5
- This leaves us with the digits 4, 6, 7, and 8 to fill the remaining two places (hundreds and tens).

3. **Determine the Remaining Digits**:
- Available digits after fixing the first and last digits: 4, 6, 7, and 8 (4 digits remaining).

4. **Calculate the Number of Combinations**:
- We need to fill the hundreds and tens places with the remaining 4 digits.
- The number of ways to choose 2 digits from the 4 available digits (4, 6, 7, 8) and arrange them is calculated using permutations:
- The number of ways to choose and arrange 2 digits from 4 is given by \( P(4, 2) \).
- This can be calculated as:
\[ P(4, 2) = 4! / (4-2)! = 4! / 2! = (4 \times 3) = 12 \]

5. **Conclusion**:
- Therefore, the total number of 4-digit numbers that can be formed under the given conditions is **12**.

### Final Answer:
The total number of numbers divisible by 5 and lying between 3000 and 4000 that can be formed using the digits 3, 4, 5, 6, 7, and 8 (without repetition) is **12**.