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.

  • 2
    $\begingroup$ That idea is correct, you have to consider both cases and add the number of possibilities for both together to get the answer $\endgroup$
    – SescoMath
    Commented Sep 17, 2018 at 13:23

2 Answers 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.

  • $\begingroup$ And you got that answer blazing fast! $\endgroup$ Commented Sep 17, 2018 at 15:21

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:

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

  • $\begingroup$ I would stress the fact that the number is even means we select the units digit first to ensure it is even, then pick the thousands digit since we still have to deal with the restriction that the leading digit cannot be zero. $\endgroup$ Commented Sep 17, 2018 at 13:55
  • $\begingroup$ Yes, I meant it like that. But I got lost in counting which lead to editing my post several times. $\endgroup$
    – Cornman
    Commented Sep 17, 2018 at 13:57

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