Trying to iron out the kinks from my thought process. I want to start backwards at the fourth digit. This gives me four options: ${0,2,4,6}$. So, _ * _ * _ * $4$.
Here's where I might be getting confused. Now I go back to the front to select the first digit. I can pick from six options which are $1,2,3,4,5,6$. However, does this leave me with six options if a $0$ is chosen for digit four, or five options if anything other than $0$ is chosen for digit four?
If this is an unsafe thought process, then call me out.
Best Answer
You consider two cases, ending in $0$ or ending in $2,4,6$.
Suppose the last digit is $0$, then as you say, you have $6$ options for the first number, $5$ for the second, and $4$ for the third. Thus, there are $120$ distinct numbers that end in $0$.
Now suppose you pick a last digit that is not $0$. There are $3$ options for this. Then, you have only $5$ numbers to pick from for the first digit. However, you still have $0$ as an option for the second and third digits. In other words, there are also $5$ options for the second digit and $4$ options for the third. So there are $300$ options in this case.
Thus, there are a total $120 + 300 = 420$ numbers with the restrictions you gave.