Number of four digit numbers ending with even integers such that their digits do not repeat

Question

How many four-digit even positive integers are there without a repetition of the digits?

My attempt:
First,we will classify such required arrangements into two groups:
$1$. Those that do contain $0$ as the end digit
To calculate this, we see that there are $9$ ways to fill the first place, $8$ ways to fill the second(we can't use $0$ and the digit at first place ), $7$ to fill the third place. Thus, there are $8\times 9\times 7$ to create required arrangements of form $1$.
$2$. Those that do not contain $0$ as the end digit.
Suppose we choose $i$ as the even number with $i\in {2,4,6,8}$. Then, we have $8$ options each for filling the first place and second place and $7$ options to fill the third. This gives us $8\times8\times 7$ ways for $2$. For each of the $4$ even end-digits, we get in total $4\times8\times8\times 7$ ways to do so.

Now if we add the final results of $1$ and $2$ we get $ 4\times8\times8\times 7+8\times 9\times 7=2296$ ways to obtain the given arrangements. But my book's answer is $4500$. What is wrong with my calculation? Please let me know.

Answer 1

Your working is correct if the repetition of digits is not allowed. $4500$ is the count of $4$ digit even numbers with repetition allowed. We know there are $9000$ numbers between $1000$ and $9999$. Half of them are even.

An alternate working if the repetition of digits is not allowed -

If we allow the thousand place to have zero, we have total of $10 \cdot 9 \cdot 8 \cdot 7 = 5040$ numbers and half of them ($2520$) will be even numbers.

We now subtract count of even numbers with $0$ in thousand place as they are not truly four digit numbers - there are $4$ choices for one's place for it to be even number ($2, 4, 6, 8$) and then that gives $8 \cdot 7 = 56$ choices for the ten's and hundred's place.

That leads to $2520 - 4 \cdot 56 = 2296$ even numbers.