How many numbers lying between 3000 and 4000 and which are divisible by 5 can be made with the digits 3, 4, 5, 6, 7 and 8? (Digits are not to be repeated in any number.)

Options

A. 11
B. 12
C. 13
D. 14

Answer

To solve the problem of how many numbers lying between 3000 and 4000 that are divisible by 5 can be made with the digits 3, 4, 5, 6, 7, and 8 (without repeating any digits), we can follow these steps:

### Step 1: Identify the range and conditions
We need to find 4-digit numbers that:
- Are between 3000 and 4000
- Are divisible by 5
- Use the digits 3, 4, 5, 6, 7, and 8 without repetition

### Step 2: Determine the first digit
Since the number must be between 3000 and 4000, the first digit must be **3**.

### Step 3: Determine the last digit for divisibility by 5
A number is divisible by 5 if its last digit is either **0** or **5**. Since we only have the digits 3, 4, 5, 6, 7, and 8 available, the last digit must be **5** (0 is not an option).

### Step 4: Fix the first and last digits
Now we have:
- First digit: 3
- Last digit: 5

This leaves us with the digits **4, 6, 7, and 8** available for the middle two digits.

### Step 5: Choose the middle digits
We need to fill the two middle positions with the remaining digits (4, 6, 7, and 8).

### Step 6: Calculate the combinations for the middle digits
We have 4 choices for the second digit and then 3 choices for the third digit (since we cannot repeat digits).

So, the number of ways to choose the middle two digits is:
- 4 choices for the second digit
- 3 choices for the third digit

Thus, the total combinations for the middle digits = \(4 \times 3 = 12\).

### Conclusion
Therefore, the total number of 4-digit numbers between 3000 and 4000 that are divisible by 5 and can be formed using the digits 3, 4, 5, 6, 7, and 8 without repetition is **12**.