Number of ways a 4 digit number can be formed using the given digits which are divisible by 2 and 5

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$\text{S} = \{0,2,4,6,8\}$. If four numbers are selected from $\text{S}$ and a four digit number ABCD is formed, then the number of such numbers which are divisible by $2$ and $5$ (all digits are not different) is?

My Attempt:

Since the numbers so formed are divisible by $2$ and $5$, they will be divisible by $10$. That means the last digit must be $0$.

Now, for the first digit, zero cannot be taken cause the digit so formed will not be a $4$ digit number. Therefore I have $4$ choices.

For the other two, I can choose anyone from the given $5$ digits.

So, my final answer should be $4 \cdot 5 \cdot 5 = 100$

However, this is a wrong answer.

The correct answer given in my book is $76$.

Any help would be appreciated.

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They want to make sure that not all digits are different. Thus, out of the $100$ you calculated, you need to subtract those where the digits are all different, of which there are $4 \cdot 3 \cdot 2 = 24$, leaving you with $76$