From digits {0, 1, 2, 3, 4, 5, 6}, how many four-digit even numbers with distinct digits can be constructed?

Question

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.

Answer 1

Your ideas are good. You might distinct into two cases. When the last digit is $0$ and when it is not $0$.

Since the number is supposed to be an even number with four digits, the last number has to be even and the first digit can not be $0$.

If the last number is $0$, there are $6\cdot 5\cdot 4$ possibilities for the other three digits.

If the last number is not $0$, the first number can not be $0$. This leaves $5$ possibilites for the first digit. Since the $0$ is illegal. And $3$ for the last digit. For the other two we have $5\cdot 4$ possibilities to choose. Since we have to again consider the $0$.

Hence $5\cdot 5\cdot 4\cdot 3$ possibilites over all.

We have to add these to get the number of possible ways to construct such a number, which should be:

$5\cdot 5\cdot 4\cdot 3+6\cdot 5\cdot 4$

Answer 2

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.