The number of six digit numbers that can be formed from the digits `1,2,3,4,5,6 & 7` so that digits do not repeat and terminal digits are even is:

Options

A. `5.4!.3!`
B. 72
C. 720
D. `5!.3!`

Answer

To solve the problem of finding the number of six-digit numbers that can be formed from the digits 1, 2, 3, 4, 5, 6, and 7, with the conditions that digits do not repeat and the terminal digits (the first and last digits) are even, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Even Digits**:
The even digits from the set {1, 2, 3, 4, 5, 6, 7} are 2, 4, and 6. Thus, we have three even digits available for the terminal positions.

**Hint**: List out the digits and identify which ones are even.

2. **Choose the First and Last Digits**:
We can choose any one of the three even digits for the first position (the first terminal digit). After selecting the first digit, we will have two remaining even digits to choose from for the last position (the last terminal digit).

- Choices for the first digit: 3 (2, 4, or 6)
- Choices for the last digit: 2 (after choosing the first)

**Hint**: Remember that once you choose a digit for the first position, it cannot be used again for the last position.

3. **Calculate the Number of Ways to Choose Terminal Digits**:
The total number of ways to choose the first and last digits is:
\[ 3 \text{ (choices for first)} \times 2 \text{ (choices for last)} = 6 \text{ ways} \]

**Hint**: Multiply the number of choices for the first and last digits to find the total combinations for these positions.

4. **Fill the Middle Four Digits**:
After selecting the first and last digits, we will have 5 digits left (since two have been used). We need to fill the middle four positions with these remaining digits.

The number of ways to arrange 5 digits in 4 positions is calculated as follows:
- For the first middle position, we have 5 choices.
- For the second middle position, we have 4 choices (one digit has been used).
- For the third middle position, we have 3 choices.
- For the fourth middle position, we have 2 choices.

The total number of arrangements for the middle four digits is:
\[ 5 \times 4 \times 3 \times 2 = 120 \text{ ways} \]

**Hint**: Use the multiplication principle for counting arrangements.

5. **Calculate the Total Number of Six-Digit Numbers**:
Now, we combine the choices for the terminal digits with the arrangements of the middle digits:
\[ \text{Total} = (\text{Ways to choose terminal digits}) \times (\text{Ways to arrange middle digits}) = 6 \times 120 = 720 \]

**Hint**: Always ensure to multiply the number of ways from different parts of the problem to get the final count.

### Final Answer:
The total number of six-digit numbers that can be formed under the given conditions is **720**.