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?
 A: The number of even four-digit decimal numbers with no digit repeated is not equal to the number of odd four-digit decimal numbers with no digit repeated.  So it's not as straightforward as diving by $2$.
I suggest you try starting with the possible values $x\ \_\ \_\ y$, since once $x$ and $y$ are determined, we can always fill in the other cells in $8 \times 7$ ways.  You'll also see why the numbers above are different (there's more pairs $(x,y)$ for which $y$ is even then for which $y$ is odd).
A: I agree with Rebecca, the number of numbers (even and odd) are not same...
My approach: four digit numbers ending with 0,2,4,6,8 with no digits repeated. 

Case #base
  1. ending with 0 : _ _ _ 0 = 9*8*7*1 = 504
      leftmost digit can be in {1,2,...9} and without repetition, each time one number is removed.
  
   Case #other
    as the number can start with any digit, except 0...

  
*ending with 2 : _ _ _ 2 = 8*8*7*1 = 448
  
*ending with 4 : _ _ _ 4 = 8*8*7*1 = 448
  
*ending with 6 : _ _ _ 6 = 8*8*7*1 = 448
  
*ending with 8 : _ _ _ 8 = 8*8*7*1 = 448

And the answer is $504+(4*448)$ =  2296
thanks ^_^
A: $4 \cdot 9  \cdot 8  \cdot 7=2016$
Since the last digti you have only $\ 2,\ 4,\ 6,\ 8\ $ four numbers to choose.
