To find how many even numbers can be formed using the digits 2, 3, 4, 5, and 6 without repeating any digit, we need to follow these steps:

1. The number must end in an even digit. Since the number has to be even, its last digit must be chosen from the even digits available: 2, 4, or 6. So, we have 3 options for the last digit.

2. Choosing the first four digits. After choosing the last digit, we have 4 remaining digits to choose from (since no digit can be repeated). The number of ways to arrange these four digits is simply the number of permutations of 4 digits.

   The number of permutations of 4 digits is given by $4! = 4 \times 3 \times 2 \times 1 = 24$.

3. Total number of even numbers. Since we have 3 choices for the last digit and 24 ways to arrange the remaining digits, the total number of even numbers is:

   $3 \times 24 = 72$

Thus, 72 even numbers can be formed using the digits 2, 3, 4, 5, and 6 without repetition.

Would you like more details or have any questions?

Here are 5 related questions for further exploration:

1. How many odd numbers can be formed using the same digits?
2. What if repetition of digits was allowed?
3. How many numbers in total can be formed using all the digits?
4. How many numbers divisible by 5 can be formed using these digits?
5. How many numbers greater than 50000 can be formed using these digits?

Tip: Factorials are fundamental in counting permutations and combinations!