We need to find the count of 6 digit numbers that are divisible by 4, with a constraint that no digit repeats.
Divisibility rule of 4 says that, for a number $x$ to be divisible by 4, last two digits of $x$ should be divisible by 4.
All the numbers having last two digits in $\{00, 04, 08, …, 92, 96\}$ would be divisible by 4. But our constraint is that none of the digits should repeat.
Desired set would become $D = \{00, 04, 08, …, 92, 96\} \ – \ \{00, 44, 88\} = \{04, 08, 12, …, 84, 92, 96\}$
$|D| = 22$
Lets’ partition $D$ into $D_{0}$ and $D_{\sim 0}$, where $D_{0}$ is the partition consisting of pair of digits having $0$ and $D_{\sim 0}$ partition not having zero in the pair.
$|D_0| = |\{04, 08, 20, 40, 60, 80\}| = 6$ , $|D_{\sim 0}| = |\{12, 16, 24, …, 84, 92, 96\}|=22-6=16$
In $D_0$, $X0$ and $0X$ format of numbers are not in equal ratio, hence we cant’ form one case for them, we need to break them down to $D_{X0} = \{20,40,60,80\}$ and $D_{0X} = \{04, 08\}$.
Similarly, we cant’ form $D_{\sim 0}$ as one case, this is because, when we take last digit in $\{2,6\}$ then second last digit has 5 possibilities $\{1,3,5,7,9\}$. Lets’ call this set $D_{OE}$.
On the Other hand, if we take last digit as $\{4,8\}$ then second last digit has $\{2,4,6,8\}$ possibilities, in total $2*4-2 = 6$ possibilities. Lets’ call this set $D_{EE}$
$$\begin{align}\text{Total count} &= D_{X0} + D_{0X} + D_{EE} + D_{OE} \cr &= \quad 8 \times 7 \times 6 \times 5 \times 4 \times 1 \cr &\quad + \ \ 8 \times 7 \times 6 \times 5 \times 1 \times 2 \cr &\quad + \ \ 7 \times 7 \times 6 \times 5 \times 6 \cr &\quad + \ \ 7 \times 7 \times 6 \times 5 \times 5 \times 2 \cr &= \boxed{33,600} \end{align}$$
