How many even four-digit numbers can be formed from the digits $0, 1, 2, 5, 6,$ and $9$ if each digit can be used only once

combinatorics

First, I know this was posted first here but I reposted because the answer wasn't clear for me and the question is 8 years old so it's not getting any activity.

I don't understand the reason of the split, I first solved the problem in the following way:

Last digit: we have $3$ options: $0,2,6$ since we need the number to be even.

First digit: we have $4$ options: all minus $0$ and digit picked previously

Third digit: we have $4$ options: all minus the one picked in the previous $2$ digits.

Second digit: we have $3$ options: all minus previous $3$ digits.

so we get $3*3*4*4$

However I have a logical problem here, if I start analyzing in a different order (for example last digit to first) then it seems like I am getting different result (example:last digit has $3$ options and third one has $5$, etc…)

Moreover, I don't understand the solution and why they split it into these $2$ cases.

Best Answer

If we chose $0$ as the last digit, "the digit $0$" and "the digit picked previously" are the same digit, meaning that in this case, you have $5$ options for the first digit.

This is the reason for the split.

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