How many 3 digit even numbers are there(No Repetition)? First find numbers ending with 0
So, 1's place-1 10's place-9 100's place-7 (2 digits are already consumed and 0 can't be used)
So 7*9*1.Im i doing the right thing?
 A: First find the even numbers that are ending up in zero
so, for no. ending up with zero are
zero at one's place so 1 combination
now 9 at hundreds and 8 at tens
Now even numbers not ending with zero i.e. ending  in 2,4,6,8
so 4 at one's place only one is used so 8 at hundreds place since 0 can not be used and one number is 0 and now 8 are left for the tens place
so, total no. of digit sequences would be
=>9*8*1+8*8*4=72+256=328
When zero is used up there is no problem for using up tens or hundreds but when zero is not used up after putting one of 2,4,6,8 numbers at one's place we have to check if 0 is ending up in the most significant place or not. To avoid that we start with hundredth place. So we ensure this first.
A: Here we consider numbers of the form xyz, where each of x, y, z represents a digit under the given restrictions. Since xyz has to be even, z has to be 0, 2, 3, 4, 6, or 8.
If z is 0, then x has 9 choices. 
If z is 2, 4, 6 or 8 (4 choices) then x has 8 choices. (Note that x cannot be zero)
Therefore, z and x can be chosen in (1 × 9) + (4 × 8) = 41 ways. For each of these ways, y can be chosen in 8 ways. 
Hence, the desired number is 41 × 8 = 328 numbers 3-digit even numbers exist with no repetitions.
A: Units digit can be among 0,2,4,6,8.
CASE A: If 0 is at units place, No. of terms possible is 1x9x8 = 72
CASE B: If 0 is not at units place but at hundred's place No of terms possible is 4x1(0)x8 = 32
CASE C: If 0 is not there at all, No of terms possible is: 4x8x7 = 224
total no's = 224+32+72 = 328
A: \begin{align}
  & \underline{\text{case 1}}\text{ : All possible even 3-digit   numbers  are given by =  {${5*9*8 =360}$} } \\  
 & \underline{\text{case 2}}\text{ : All even number with ''0'' at }hundreds\text{ }place\text{ are given by =  {${4*8*1=32}$} } \\
 & \underline{The\text{ number of even 3 digit is = case1 -case 2 =  {360-32=328}} } 
\end{align}
A: Total three digits numers are 900
We have total 10 digits which are 0,1,2,3,4,5,6,7,8,9
But a leading zero does not have any value so we cant have zero at 100th place
So for 100th place we have 9 digits, for tenth place we have 10 digits and for units place we have 10 digits
So total 3 digits number are 
9*10*10
=900
If we talk about even numbers then we have only 5 digits fot units place 0,2,4,6,8
So total 3 digits even numbers would be 9*10*5
=450
A: There are some good answers here. But I just want to mention one more approach:

*

*First calculate total 3 digit numbers without repetition: 9 x 9 x 8 = 648

*Then calculate 3 digit odd numbers without repetition: 8 x 8 x 5 = 320

*Now subtract the number of odd numbers from total: 648 - 320 = 328
A: The total 3 digit numbers are 999 including preceding zeros and there are 999/2 even numbers.
so total three digit even numbers are 499.
A: On the number line the 3 digit numbers are : 100 - 999
So if I would start at 1 - 999 then I have 999 numbers in total.
from those I will take away the one digit numbers : 9
and the two digit numbers: 90
I.e: 999-99 = 900 3 digit numbers.
Now to count the even numbers: we start at 100 and skip count by 2.
Or perhaps we can divide by 2: 900/2 = 450 even numbers.
Please let me know if you agree.
