How many different even 4-digit numbers can be created if digits 0-6 can be used and the first digit is nonzero. Repetition is not allowed.

Question

I know that the first digit will be between 1-6 because 0 cannot be chosen but I do not know how to proceed further. How do you solve this?

Answer 1

Case 1: The first digit is even (2, 4, or 6)

In this case, the last digit must be some other even digit, while the middle two digits must be different from each other and from the first and last digits. This gives $3 \cdot 5 \cdot 4 \cdot 3=180$ 4-digit even numbers with distinct digits less than 7 where the first digit is even.

Case 2: The first digit is odd (1, 3, or 5)

In this case, the last digit can be any even digit, while the middle two digits must still be different from each other and from the first and last digits. This gives $3 \cdot 5 \cdot 4 \cdot 4=240$ 4-digit even numbers with distinct digits less than 7 where the first digit is odd.

Adding both cases together, the total number of 4-digit even numbers with distinct digits less than 7 is $180+240=420$.

Answer 2

You only need to be careful in picking the first digit and the last digit (to ensure your number is even).

Split this in as many cases as you need and then tackle it! For example, start by picking the first digit in 1-6, like you said. Then, how can you pick the last digit?