From the digits 1, 2, 3, 4, 5, 6 how many three - digit odd numbers can be formed when the repetition of the digits is not allowed ?

Answer

To solve the problem of finding how many three-digit odd numbers can be formed from the digits 1, 2, 3, 4, 5, 6 without repetition, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Odd Digits**:
The odd digits available from the set {1, 2, 3, 4, 5, 6} are 1, 3, and 5. Therefore, we have 3 odd digits.

2. **Choose the Ones Place**:
Since we want to form an odd number, the ones place must be filled with one of the odd digits. We have 3 choices for the ones place: 1, 3, or 5.

3. **Choose the Hundreds Place**:
After selecting a digit for the ones place, we have to choose a digit for the hundreds place. Since repetition is not allowed, we can choose from the remaining 5 digits (the original 6 digits minus the one used in the ones place).

4. **Choose the Tens Place**:
After filling the hundreds and ones places, we will have 4 digits left to choose from for the tens place.

5. **Calculate the Total Combinations**:
The total number of three-digit odd numbers can be calculated by multiplying the number of choices for each place:
- Choices for the ones place: 3 (1, 3, or 5)
- Choices for the hundreds place: 5 (remaining digits after choosing the ones place)
- Choices for the tens place: 4 (remaining digits after choosing the hundreds and ones places)

Therefore, the total number of three-digit odd numbers is:
\[ \text{Total} = (\text{Choices for ones}) \times (\text{Choices for hundreds}) \times (\text{Choices for tens}) = 3 \times 5 \times 4 \]

6. **Perform the Calculation**:
\[ 3 \times 5 = 15 \]
\[ 15 \times 4 = 60 \]

Thus, the total number of three-digit odd numbers that can be formed is **60**.