Rearranging digits for multiples of 4

Question

How many ways are there to arrange the digits $1,2,3,4,5,6,7$ so that a $7$ digit number is formed, such that the number is a multiple of $4$.

I know that when the second last digit is an odd number the last number must be "$2$" or "$6$". Additionally, I know that when the second last digit is an even number the last number must be "$0$", "$4$", or "$8$".

I approached the problem in cases:

Case $1$: Second last digit odd number $= 5!(4)(2)$
Case $2$: Second last digit even number "not $4$" $= 5!(2)(3)$
Case $3$: Second last digit even number "$4$" $= 5!(2)$

However, this is not the correct answer. Their answer is $5!(2) + 5!(2)(4)$.

Can anyone explain to me where I went wrong?

Answer 1

You got the number of options for Case 1 correct, which is indeed $5! \cdot 4 \cdot 2$

For case 2, note that you don't have $8$ or $0$. So, when the last two digits are even, you only have $2$ options: it ends in $24$ or $64$. Hence you have $5! \cdot 2$ options for Case 2

Otherwise, good thinking and good work!

Answer 2

In order for the $7$-digit number to be divided by $4$ ,the last $2$ digits of that number has to be divided by $4$(or has to be $00$ but we don't have $0$ among the numbers thus we skip this possibility).
So, you could only have $12, 16, 24,32, 36, 52,56, 64, 72, 76$ as the last two digits. In total you have $10$ possible choices for the last two digits.
Let's say the last two digits is $12$, so the number of ways you can get is $5!$
You can do that for other choices like $16,24...$ etc
As you consider all the other choices too, you end up with $10\cdot5!$

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But to do it by the way you do it:
Case $1$ is okay.
Case $2$: is okay too.
Case $3$: Second last digit even number "$4$". There is no way to put any number in the list next to the $4 $ so that you can eventually have a $7$-digit number that is divided by $4$. So this case counts $0$ way.

So, again we end up with $5!\cdot 8+5!\cdot 2+0=5!\cdot 10$