I am currently working on this problem:
How many 4-digit multiples of 2 use only digits 0 through 4, with no digits repeating?
I understand/know the basic idea of how to solve this kind of problem:
You multiply the number of possible numbers that could be in each place, for example, for How many 4 digit numbers use only digits 0 through 3, with no digits repeating?, the answer would be 3 * 3 * 2 * 1, or 18.
In my problem, the last digit of any even number is either 0, 2, or 4. That gives us this current multiplication equation:
_ * _ * _ * 3
The reason I am stuck on this problem is because, in the first space, there could either be 4 possibilities(1, 2, 3, 4) or 3 possibilities(1, 3, 4 OR 1, 2, 3), depending on if 2 or 4 is used in the ones column. And depending on the 1st column, the following numbers of possibilities will be one minus the last number.
Now, because there can be two possible number of possibilities in the left-most/4th space, there can be two multiplication problems, which makes this problem confusing for me:
4 * 3 * 2 * 3
3 * 2 * 1 * 3
So my question is: How would you solve this problem?