Final answer:
There are 360 possible even four-digit numbers that can be formed with the digits 1, 2, 3, 4, 5, 6, and 7 without repetition, by considering the combinatorial possibilities for each digit place while ensuring the last digit is even.
Step-by-step explanation:
To answer the question of how many even numbers can be formed using the digits 1, 2, 3, 4, 5, 6, and 7 without repetition, we need to consider the properties of even numbers. An even number must end in either 2, 4, or 6. We can use combinatorial counting to calculate the possibilities.
For a four-digit number to be even, its last digit must be 2, 4, or 6. We have 3 choices for the last digit. For the first digit, we can't have 0 and we cannot use the digit chosen as the last digit, so we have 6 possibilities. For the second digit, we can choose from the remaining 5 digits, and for the third digit, we have 4 remaining choices. Therefore, the total number of even four-digit numbers we can form is 3 (choices for last digit) × 6 (choices for first digit) × 5 (choices for second digit) × 4 (choices for third digit).
Calculating this, we have:
- 3 × 6 = 18
- 18 × 5 = 90
- 90 × 4 = 360
So, there are 360 possible even four-digit numbers that can be formed with the digits 1, 2, 3, 4, 5, 6, and 7 without repetition.