Consider the number of $3$ distinct numbers formed with the digits $2,3,5,8,9$. How many of them are even? I'm trying to answer the following:

Consider the number of $3$ distinct numbers formed with the digits $2,3,5,8,9$. How many of them are even?

I first tried to make the following counting: First choose one of the even digits for the end, that's two choices. Then choose the rest of them at will. This yields $2*3*2=12$. But the answer is not $12$, it need to be made like: $1*4*3+1*4*3=24$. I don't get what is wrong with the first counting, can you help me?
 A: Well you start from the last digit so there are 2 ways to choose it, but you only used one of them so you have 4 numbers left. So you can just do 4 times 3 for the remaining digits. So you would get 4*3*2 which is 24.
A: The ones digit can have one of two options; the tens digit can have one of four, and the hundreds digit can have one of three: 
$$2 \cdot 3 \cdot 4 = 24$$ different possibilities.
The reason your original reasoning didn't work out is when you assigned two different possibilities to the ones digit, and then discounted the fact that the other of $\{2,8\}$ that you didn't select for the ones digit can still be in the tens digit. Once you fix that issue, it should be easy to see the solution. 
A: The only 2 digits that make it even are 2 and 8.
Case 1 (2):
There are 2 digits that can be changed, choosing from the set {3,5,8,9}
Therefore the probability is $_4P_2=12.$
Case 2 (8):
There are 2 digits that can be changed, choosing from the set {3,5,2,9}
Therefore the probability is $_4P_2=12.$
The number of successful cases is 12+12=24.
A: There are $\binom{2}{1}$ ways to select the last digit out of the numbers $2$ and $8$, and $\binom{4}{2}2!$ ways to select and to permute the remaining $2$ digits out of the remaining $4$ digits available. Hence,
$$\binom{2}{1}\binom{4}{2}2! = 24$$
