How many odd numbers greater than 80000 can be formed using the digits 2, 3, 4, 5 and 8 if each digit is used only once in a number ?

Answer

To solve the problem of how many odd numbers greater than 80,000 can be formed using the digits 2, 3, 4, 5, and 8 with each digit used only once, we can follow these steps:

### Step 1: Identify the conditions
We need to form a 5-digit number that is:
1. Greater than 80,000
2. An odd number

### Step 2: Determine the first digit
Since the number must be greater than 80,000, the first digit can only be 8 (the only digit that meets this condition from the given digits: 2, 3, 4, 5, 8).

### Step 3: Determine the last digit
To ensure the number is odd, the last digit must be either 3 or 5 (the only odd digits available from the given set).

### Step 4: Calculate the possibilities for each case
#### Case 1: Last digit is 3
- The first digit is 8.
- The last digit is 3.
- The remaining digits to fill the middle three positions are 2, 4, and 5.

The number of ways to arrange the remaining three digits (2, 4, 5) is:
\[ 3! = 6 \]

#### Case 2: Last digit is 5
- The first digit is 8.
- The last digit is 5.
- The remaining digits to fill the middle three positions are 2, 3, and 4.

The number of ways to arrange the remaining three digits (2, 3, 4) is:
\[ 3! = 6 \]

### Step 5: Total combinations
Now, we add the combinations from both cases:
- From Case 1 (last digit 3): 6 combinations
- From Case 2 (last digit 5): 6 combinations

Thus, the total number of odd numbers greater than 80,000 is:
\[ 6 + 6 = 12 \]

### Final Answer
The total number of odd numbers greater than 80,000 that can be formed using the digits 2, 3, 4, 5, and 8 is **12**.

---