How many number of 5 digit even number can be formed by using the digits 0,1,2,3,4,5,6 and 7 while
repetition not allowed.

To find how many 5-digit even numbers can be formed using the digits \( 0, 1, 2, 3, 4, 5, 6, 7 \) without repetition, we need to break the problem down step by step.

### Key considerations:
1. **Even number condition**: For a number to be even, the last digit must be one of the even digits: \( 0, 2, 4, 6 \).
2. **5-digit number**: The first digit cannot be zero (since that would make it a 4-digit number).

Let's solve this step by step.

### Step 1: Choose the last digit
The last digit must be one of the even digits \( 0, 2, 4, 6 \), so there are 4 choices for the last digit.

### Step 2: Choose the first digit
The first digit must be non-zero (so it has to be from \( 1, 2, 3, 4, 5, 6, 7 \)). If 0 is already used as the last digit, we have 7 choices for the first digit. Otherwise, if the last digit is one of \( 2, 4, 6 \), then we have 6 remaining choices for the first digit (since that even digit is already used as the last digit).

### Step 3: Choose the remaining three digits
After choosing the first and last digits, we are left with 6 digits for the second position, 5 digits for the third position, and 4 digits for the fourth position.

### Now, let's break it into two cases:

#### Case 1: The last digit is 0
- For the first digit, we can choose from \( 1, 2, 3, 4, 5, 6, 7 \), so there are 7 choices.
- For the second digit, we now have 6 remaining digits to choose from.
- For the third digit, 5 remaining digits.
- For the fourth digit, 4 remaining digits.

So, the total number of possibilities in this case is:
\[
7 \times 6 \times 5 \times 4 = 840
\]

#### Case 2: The last digit is 2, 4, or 6
- For the first digit, we can choose from \( 1, 3, 5, 7 \), and one of these numbers must be the first digit, so there are 6 choices.
- For the second digit, we now have 6 remaining digits to choose from.
- For the third digit, 5 remaining digits.
- For the fourth digit, 4 remaining digits.

So, the total number of possibilities in this case is:
\[
3 \times 6 \times 6 \times 5 \times 4 = 2160
\]

### Total possibilities
Now, we add the two cases together:
\[
840 + 2160 = 3000
\]

Therefore, the total number of 5-digit even numbers that can be formed is \( \boxed{3000} \).

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Do you want more details, or have any other questions?

Here are some related questions:
1. How would the answer change if repetition of digits were allowed?
2. What if the number had to be odd instead of even?
3. How many 6-digit even numbers could be formed under the same conditions?
4. What would the result be if we excluded certain digits, such as 7?
5. How many 5-digit numbers can be formed that are divisible by 5?

**Tip**: When solving combinatorics problems, always identify key constraints such as digit restrictions and conditions (e.g., even, odd, etc.) before calculating.

How many 5-digit even numbers can be formed by using the digits 0, 1, 2, 3, 4, 5, 6, and 7, while repetition is not allowed?

Explore how to calculate the number of 5-digit even numbers that can be formed using the digits 0, 1, 2, 3, 4, 5, 6, and 7 without repetition. This combinatorics problem involves key concepts such as permutations, even numbers, and the basic counting principle. Suitable for high school students in Grades 9-12.

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