The number of numbers divisible by 5 and lying between 40000 and 50000 that can be formed from the digits 0,3,4,5,8 and 9, when repetition of digits is not allowed is

Options

A. 431
B. 48
C. 432
D. 84

Answer

To solve the problem of finding the number of 5-digit numbers divisible by 5 and lying between 40,000 and 50,000 that can be formed from the digits 0, 3, 4, 5, 8, and 9 without repetition, we can follow these steps:

### Step 1: Identify the range and structure of the number
Since we need a 5-digit number between 40,000 and 50,000, the first digit must be **4**.

**Hint:** The first digit determines the range of the number.

### Step 2: Determine the last digit for divisibility by 5
For a number to be divisible by 5, its last digit must be either **0** or **5**.

**Hint:** Check the divisibility rules for 5 to determine the possible last digits.

### Step 3: Case Analysis based on the last digit
We will analyze two cases based on the last digit:

**Case 1:** Last digit is **0**
- The number structure is: 4 _ _ _ 0
- The remaining digits available are: 3, 5, 8, 9 (since 4 and 0 are used).
- We have 4 remaining digits, and we need to fill 3 positions.

**Case 2:** Last digit is **5**
- The number structure is: 4 _ _ _ 5
- The remaining digits available are: 0, 3, 8, 9 (since 4 and 5 are used).
- Again, we have 4 remaining digits, and we need to fill 3 positions.

### Step 4: Calculate the number of combinations for each case

**For Case 1 (Last digit is 0):**
- The first digit is fixed as 4.
- We can choose any 3 digits from the remaining 4 (3, 5, 8, 9).
- The number of ways to arrange 3 digits is given by \(3!\) (factorial of 3).

Calculating:
- Number of arrangements = \(3! = 6\)
- Total combinations for Case 1 = 6

**For Case 2 (Last digit is 5):**
- The first digit is fixed as 4.
- We can choose any 3 digits from the remaining 4 (0, 3, 8, 9).
- The number of ways to arrange 3 digits is also \(3!\).

Calculating:
- Number of arrangements = \(3! = 6\)
- Total combinations for Case 2 = 6

### Step 5: Combine the results from both cases
Now, we add the total combinations from both cases to find the final answer.

Total combinations = Combinations from Case 1 + Combinations from Case 2
Total combinations = 6 + 6 = 12

### Final Answer
The total number of 5-digit numbers divisible by 5 and lying between 40,000 and 50,000 that can be formed from the digits 0, 3, 4, 5, 8, and 9 without repetition is **12**.

---