How many four-digit even numbers can be formed from the set         {0, 1, 2, 3, 4, 5, 6, 7} if repetition of digits is NOT allowed?1 pointA. 1470B. 750C. 720D. 540

To solve this problem, we need to consider the conditions given:

1. The number must be a four-digit number.
2. The number must be even.
3. Repetition of digits is not allowed.

Let's break it down:

1. The last digit of the number must be even. Since 0 is also an even number, we have 4 choices for the last digit (0, 2, 4, 6).

2. The first digit can be any of the remaining 7 digits (since it can't be 0 or the digit we chose for the last place). So, we have 7 choices for the first digit.

3. For the second digit, we can choose from the remaining 6 digits. So, we have 6 choices for the second digit.

4. For the third digit, we can choose from the remaining 5 digits. So, we have 5 choices for the third digit.

So, the total number of four-digit even numbers that can be formed is 7 (choices for the first digit) * 6 (choices for the second digit) * 5 (choices for the third digit) * 4 (choices for the last digit) = 840.

So, the correct answer is not listed in the options given.