Counting Number of even and distinct digits The Question was: The number of even four-digit decimal numbers with no digit repeated. 
So the first digit cannot be 0 so there are 9 ways to choose a digit. Then for the 3rd, 2nd and 1st digits there would be respectively 9 ways (adding back the zero as an option), 8 ways, 7 ways. So then the total possibilities are 4,536 but then since we are looking for even numbers we divide this number by 2 and we get 2,268. I don't understand why this is wrong. Can someone please help me?

$4 \cdot 9  \cdot 8  \cdot 7=2016$
Since the last digti you have only $\ 2,\ 4,\ 6,\ 8\ $ four numbers to choose.
