TITLE: How many positive integers less than 1000 have distinct digits and are even?
QUESTION [5 upvotes]: I am not looking for an answer on this. Just need to clarify why my approach is failing -
$N_1 + N_2 + N_3$, i.e. single digit, double digit, 3 digit
single $= 2, 4, 6, 8$, i.e 4 
double = X non-zero $= 8 \cdot 4 = 32$ X zero $= 9 \cdot 1 = 9$
Now the confusing part three digit, breaking into 4 cases 
X zero zero $= 9$ 
X nz nz $= 7 \cdot 8 \cdot 4 = 224$ 
X z nz $= 8 \cdot 1 \cdot 4 = 32$ 
X nz z $= 8 \cdot 9 \cdot 1 = 72$
Three digit total comes to $= 9 + 224 + 32 + 72 = 337$. This answer is wrong and it should be $328$. What am I missing in the logic? Please suggest.

REPLY [3 votes]: (I was going to add this as a comment, but it turned out to be too long.)
Here are two alternate ways to handle the 3-digit case:  
a) If the last digit is not zero, there are 4 choices for the last digit, 8 choices for the first digit, and 8 choices for the middle digit.  
b) If the last digit is zero, there are 9 choices for the first digit and 8 choices for the middle digit.  
This gives a total of $4\cdot8\cdot8+1\cdot9\cdot8=328$.  

Alternatively, 
1) If the first digit is even, there are 4 choices for this digit, 4 choices for the last digit, and 8 choices for the middle digit.  
2) If the first digit is odd, there are 5 choices for this digit, 5 choices for the last digit, and 8 choices for the middle one.
This gives a total of $4\cdot4\cdot8+5\cdot5\cdot8=328.$