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I'm trying to find the number of 4-digits even numbers, such that no digit repeats.

This is simple, and the solution is 2296, as is explained in How many $4$ digit even numbers have all $4$ digits distinct?

I've tried solving it with complementary sets: "The number of 4-digits even numbers" minus "The number of 4-digits even numbers such that all the digits are the same"

However with the second approach I get 4996. What am I missing? can I find the solution using complementary sets?

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If you're subtracting the cases in which all four digits are the same you are not removing all of the cases which contain a repeat. You also need to remove the cases in which $2$ digits are the same and $3$ digits are the same.

To get the answer desired (that is, to find the complement of the original answer which was how many four-digit evens with no repeats) we must find all numbers in which at least one digit is the same. Thus with $9\cdot 10 \cdot 10 \cdot 5=4500$ four-digit even numbers with no restrictions we can remove the original number you found $2296$ (the number with no repeats) to find the complement which is the number of four-digit even numbers where at least one digit is a repeat $2204$.

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The fourth and last digit has 5 even options.

The third digit has 9 options. (0-9 except the last digit)

The second digit has 9 options. (0-9 except the third digit)

The first digit has 8 options, unless the second digit is 0, then it has 9 options.

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