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

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.

• $4500$ is the number of even numbers from $1000$ through to $9999$. Some of those will have repetition of digits in some sense, such as $8888$ Jul 6, 2021 at 10:27
• You reversed your descriptions of cases 1 and 2. You should have $9 \cdot 8 \cdot 7 \cdot 1$ four-digit positive even integers with distinct digits with units digit $0$ and $8 \cdot 8 \cdot 7 \cdot 4$ four-digit positive even integers with distinct digits with units digit not equal to zero. Jul 6, 2021 at 11:14
• Oh sorry.I will edit it now. Thankyou. Jul 6, 2021 at 11:23

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.
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.